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What are you here for?
14Oscillations
A 0.500 kg block sits on a smooth bench, held against a spring that has been pulled 15.0 cm sideways. Let go, and everything you have learned so far gives its speed at the middle: 3.00 m/s, in one line, from the stored energy. Now answer the question the demonstrator actually asks, which is when it gets there. The obvious calculation gives 0.0707 s and it is wrong by ten per cent, for every spring and every mass, and the size of that error never changes.
By the end of this section you can write down where the block is at any instant, produce the 0.0785 s that the obvious method misses, say in one sentence why the miss is always ten per cent, and do the same for a swinging pendulum and for an oscillator that is slowly dying away.
In 60 seconds
When the net force on a body is proportional to how far it sits from one place and points back towards it, the motion is a cosine in time whose rate is fixed by the stiffness and the mass alone, so the is the one thing you can quote before knowing how hard anybody pulled.
The condition that defines the whole section
$$F = -kx \quad\Longleftrightarrow\quad a = -\frac{k}{m}\,x$$
checking whether a problem is a harmonic one at all; $x$ is measured from the and nowhere else
Position at every instant
$$x(t) = A\cos(\omega t + \varphi)$$
the question names a time, an instant, a or a number of cycles
, period and of a mass on a spring
$$\omega = \sqrt{\frac{k}{m}}, \qquad T = \frac{2\pi}{\omega} = 2\pi\sqrt{\frac{m}{k}}, \qquad f = \frac{1}{T}$$
always, as the first line of any spring problem; note that the is absent
Velocity and acceleration
$$v = -A\omega\sin(\omega t + \varphi), \qquad a = -\omega^{2}x$$
you need a speed or a force at a named instant rather than at a named position
Speed at a given position, without any time
$$v = \pm\,\omega\sqrt{A^{2} - x^{2}}$$
the question links a speed to a position and never mentions a clock
the amplitude visibly shrinks; for light damping the frequency is barely touched
Three most common mistakes
Believing that pulling the block twice as far makes the trip twice as long. It does not make it any longer at all: the period contains the stiffness and the mass and nothing else. Everything else about the motion doubles, and the clock does not move.
Feeding degrees into the cosine. The argument of $\cos(\omega t + \varphi)$ is an angle in radians because $\omega$ is in radians per second; a calculator left in degree mode turns a time of 0.110 s into 6.32 s and the answer is out by a factor of 57.
Confusing the two maxima: $v_{\max} = A\omega$ and $a_{\max} = A\omega^{2}$. They differ by one factor of $\omega$, which for a stiff spring is a factor of twenty, and the units are the fastest way to catch it.
The assessment table gives 20% to each midterm, 25% to the final, 10% to quizzes in total, 5% to homework and 20% to the laboratory. It says nothing about which topic sits on which paper, so nothing is claimed here about that. What can be said is structural: this section is built on the spring energy of the energy sections and on the moment of inertia of the rotational ones, so a weakness there shows up here twice.
How much time do you have?
10 minutes
You leave with the two lines that answer most spring questions, $\omega = \sqrt{k/m}$ and $x = A\cos(\omega t + \varphi)$, with the pendulum period beside them and the three errors that cost the most marks.
In 60 seconds · Formula card · Simple harmonic motion: the position at every instant · The pendulum: gravity making a spring out of a string · Mistake ledger
45 minutes
You add the parts that separate a pass from a good mark: getting the amplitude and the out of what the question tells you at $t=0$, the energy route that answers a speed question without any clock at all, and the fading ladder where you write the reasoning yourself.
In 60 seconds · Conventions used here · The restoring force: why anything repeats at all · Simple harmonic motion: the position at every instant · Velocity and acceleration: the same curve, shifted · Energy in an oscillation: one number traded back and forth · The pendulum: gravity making a spring out of a string · Choosing between the energy route and the time route · Fading ladder · Exam level worked example · Mistake ledger
full reading
Everything above plus the two blocks that make the subject stop being a formula sheet: why the cosine appears at all, which is the shadow of a turning point, and what happens to a real oscillator that loses energy and is then pushed at the wrong rate.
In 60 seconds · Conventions used here · What you need first · The restoring force: why anything repeats at all · Simple harmonic motion: the position at every instant · Velocity and acceleration: the same curve, shifted · Energy in an oscillation: one number traded back and forth · Where the cosine comes from: the shadow of a turning point · The pendulum: gravity making a spring out of a string · Real oscillators: damping and resonance · Method boxes · Contrast pairs · Fading ladder · Exam level worked example · Practice A to D · Mistake ledger · Formula card · Glossary · Check yourself
By the end of this section
Identify the equilibrium position of a system, show that the net force on it takes the form $-kx$ for a displacement $x$ from that position, and read off the .
Write the position of an oscillator at every instant as $x = A\cos(\omega t + \varphi)$, compute the period and the frequency from the stiffness and the mass, and fix $A$ and $\varphi$ from the position and the velocity at one instant.
Compute the velocity and the acceleration of an oscillator at any instant or at any position, and say where each of them is largest and where each is zero.
Apply the constant total $\tfrac12 kA^{2}$ to get a speed from a position or a position from a speed without using a clock, and split the total into its kinetic and stored parts at any point.
Explain the cosine as the shadow of a point moving steadily around a circle, and use that picture to read the meaning of the amplitude, the angular frequency and the phase constant off a diagram.
Derive the period of a simple pendulum from the tangential component of the weight, state the small angle condition with the size of the error it costs, and extend the result to a rigid body swinging about a pivot.
Describe how an oscillator that loses energy decays, compute how many cycles it survives, and locate the driving frequency at which a driven system responds most violently.
Syllabus coverage
Oscillations
The restoring force and the condition for simple harmonic motion, the position, velocity and acceleration as functions of time, the period and the frequency of a mass on a spring, the energy of the motion, the , the simple and the physical pendulum, and oscillators that are damped or driven
The week line carries no chapter numbers, so no chapter number is quoted anywhere in this section. The scope taken is the standard content of a chapter with that title in the set textbook, split across the seven blocks named above.
covered
, forced oscillation and resonance
What happens to an oscillator that loses energy, and what happens when something pushes it at a chosen rate
The week line reads Oscillations with no qualification, so the boundary between the ideal oscillator and the damped one cannot be read off it. Everything up to the pendulum is certainly inside; damping and resonance close the same textbook chapter, so they are flagged rather than folded in silently, and are developed with one worked example each.
off_syllabus
Solving the equation of motion rather than verifying a proposed solution
The general theory of the differential equation $ma = -kx$ and of its damped cousin
This section proposes the cosine, substitutes it, and argues that two free constants are exactly what a second derivative allows, which is enough to know the answer is unique. The systematic method for producing it, and the trial solution behind the damped formula, are named and not developed.
off_syllabus
Waves: an oscillation that travels
What happens when neighbouring oscillators are coupled so that the motion moves along
One sentence at the very end of the section, as a signpost only, because this is the last week of the course. Nothing about waves is used, computed or examined here.
off_syllabus
Recall first
Hooke's law and the stiffness of a spring
An ideal spring pulls back with $F = -kx$, where $x$ is the stretch or the squash measured from its natural length and $k$ is in newtons per metre. The number $k$ is a property of the spring: a stiff one has a large $k$.
It is the only force law in this section that is written down rather than derived, and every result follows from its shape.
The energy stored in a stretched spring
$U = \tfrac12 kx^{2}$, the area under the straight line of force against stretch. It does not care about the sign of $x$: a squash of 3 cm stores exactly what a stretch of 3 cm stores.
The whole energy block is this expression traded against $\tfrac12 mv^{2}$, and it also produces the speed at a position without any clock.
Conservation of mechanical energy
When only conservative forces do work, $\tfrac12 mv^{2} + U$ has the same value at every instant. On a rough surface the total falls by $f_k d$ over a distance $d$.
It gives the amplitude of an oscillation that is set up by something else, such as a block arriving with a speed, and it is what the damped block loses.
Newton's second law with acceleration as a second derivative
$F_{\rm net} = ma$ with $a = \dfrac{d^{2}x}{dt^{2}}$. Constant acceleration formulas such as $x = \tfrac12 at^{2}$ are legal only while that acceleration is a fixed number.
The reason the obvious calculation in the opening fails is that here $a$ is a different number at every point of the motion, and the second derivative is what replaces it.
Uniform circular motion
A point going round a circle of radius $r$ at a steady angular velocity $\omega$ has speed $v = \omega r$ and acceleration $a = \omega^{2} r$ pointing at the centre. Its coordinates are $x = r\cos\theta$ and $y = r\sin\theta$ with $\theta = \omega t + \varphi$.
The reference circle block turns this into a proof that the shadow of such a point is an oscillation, and the centripetal acceleration turns into a check on the maximum acceleration.
Torque, moment of inertia and the rotational second law
$\tau_{\rm net} = I\alpha$ about a fixed axis, with $\alpha = \dfrac{d^{2}\theta}{dt^{2}}$. For a uniform rod of length $L$ about one end $I = \tfrac13 ML^{2}$, about its centre $I = \tfrac{1}{12}ML^{2}$, and for a uniform disc about its centre $I = \tfrac12 MR^{2}$.
A swinging rigid body is an oscillation in an angle rather than in a position, and every step of the pendulum argument is the rotational law with the tangential weight as the torque.
The parallel axis theorem
$I = I_{\rm cm} + Md^{2}$ for an axis a distance $d$ from a parallel axis through the centre of mass.
It is the only way to get the moment of inertia of a body about the pivot it actually swings from, which is never its centre.
The small angle behaviour of the sine
For an angle measured in radians, $\sin\theta \approx \theta$, and the approximation is low by 0.13% at $5^{\circ}$, by 1.14% at $15^{\circ}$ and by 4.51% at $30^{\circ}$.
It is the single approximation that turns a pendulum into a harmonic oscillator, and the numbers above are what let you say how much you paid for it.
Try it yourself first (3 questions)
1§14.0 — spring energy launching a block●●○○○
Before any of the new material: a block is held against a squashed spring on a smooth horizontal bench and then released. Nothing here is new; it is the calculation the whole section is going to sit on top of. Getting it wrong is not a problem, it just tells you which earlier block to reread first.
Given
$m = 0.400\ \mathrm{kg}$
$k = 250\ \mathrm{N/m}$
the spring is squashed by $0.0800\ \mathrm{m}$ and the block is at rest
the bench is smooth and the block leaves the spring at the natural length
Find
(a) Find the energy stored in the squashed spring.
(b) Find the speed of the block at the instant it leaves the spring.
Hint 1/4
Nothing moves at the start and nothing is stored at the end, so the whole of one quantity turns into the whole of another. Name those two quantities before writing anything down.
Hint 2/4
The stored energy of a spring is $U = \tfrac12 kx^{2}$ and the kinetic energy is $\tfrac12 mv^{2}$; on a smooth bench the sum of the two is the same at both instants.
Hint 3/4
The numbers again: $k = 250$ N/m, squash $x = 0.0800$ m, $m = 0.400$ kg, and the block starts at rest.
Hint 4/4
The store is 0.800 J and the block leaves at 2.00 m/s.
Show solutionPut a number on the store
$$U = \tfrac12 k x^{2} = \tfrac12(250)(0.0800)^{2}$$
the squash is measured from the natural length, and the sign of it never matters because it is squared
$$U = 0.800\ \mathrm{J}$$
this is the whole budget for the trip, because the block starts at rest
Spend it all on motion
$$\tfrac12 m v^{2} = 0.800\ \mathrm{J}$$
at the natural length the spring stores nothing, so every joule has gone into the block
the bench is smooth, so nothing was taken out on the way
Answer $$\boxed{\;U = 0.800\ \mathrm{J},\qquad v = 2.00\ \mathrm{m/s}\;}$$
Check
Size check: 2.00 m/s is a brisk walking pace, which is the right size for a small block launched by a spring you could squash with one hand. A result of 20 m/s would have meant a factor of 100 lost somewhere.
Hold on to the 0.800 J. Later in this section the same block, still attached to the spring, will keep trading that number back and forth for ever, and its amplitude will be exactly the 0.0800 m it started at.
2§14.0 — does a bigger pull take longer●●○○○
A block on a smooth bench is attached to a spring. It is pulled 4.0 cm to the right and released from rest, and a stopwatch says it reaches the middle 0.12 s later. The experiment is repeated with everything the same except that the block is pulled 8.0 cm before release. This one is worth answering before you read on, because the answer is the point of the section.
Given
the same block and the same spring in both runs
first run: pulled 4.0 cm, reaches the middle after 0.12 s
second run: pulled 8.0 cm from the same equilibrium position
the bench is smooth in both runs
Find
(a) Choose the time at which the block reaches the middle in the second run.
Hint 1/4
Compare what is different between the runs and what is the same. Two things change in the second run and they change in opposite directions.
Hint 2/4
The distance to cover is doubled, and the force at the release point, and therefore the acceleration there, is also doubled because the spring is pulled twice as far.
Hint 3/4
The first run took 0.12 s over 4.0 cm. In the second run the trip is 8.0 cm and the starting acceleration is twice as large.
Hint 4/4
The two effects cancel exactly and the time is 0.12 s again.
Show solutionScale the two runs against each other
$$x \to 2x \;\Rightarrow\; F = -kx \to 2F$$
the spring is linear, so at every corresponding fraction of the trip the force is twice as big
$$a \to 2a \ \text{everywhere}$$
the mass has not changed, so the acceleration doubles at every corresponding point
Ask what that does to the clock
$$x \sim \tfrac12 a t^{2} \;\Rightarrow\; t \sim \sqrt{x/a}$$
for a fixed shape of motion, the time scales as the square root of distance over acceleration
distance and acceleration both doubled, so the ratio is untouched and so is the time
Answer $$\boxed{\;t = 0.12\ \mathrm{s},\ \text{the same as before}\;}$$
Check
Independent check from the other end: the section will show that $T = 2\pi\sqrt{m/k}$, an expression in which the amplitude does not appear at all. Two arguments, one by scaling and one by formula, and they agree.
A spring is the one force law for which a bigger swing costs no extra time, and it is exactly the linearity of $F=-kx$ that buys it. Break the linearity and the property goes.
3§14.0 — a point going round a circle●●○○○
One recall from the rotational material, because the shape of the answer to this whole section is hiding in it. A small marker is glued to the rim of a wheel that turns steadily.
the marker is at the angle $\theta = \omega t$ measured from the $x$ axis
Find
(a) Find the speed of the marker and the time for one revolution.
(b) Write the $x$ coordinate of the marker as a function of time, and state its largest value.
Hint 1/4
Two of the three things asked for come straight from a definition you already have, and the third is a piece of trigonometry about where a point on a circle sits.
Hint 2/4
For steady circular motion $v = \omega r$ and one revolution is $2\pi$ radians of angle, so $T = 2\pi/\omega$. The coordinates of a point at angle $\theta$ on a circle of radius $r$ are $x = r\cos\theta$ and $y = r\sin\theta$.
Hint 3/4
The numbers again: $r = 0.250$ m and $\omega = 4.00$ rad/s, with $\theta = \omega t$.
Hint 4/4
The speed is 1.00 m/s, one revolution takes 1.57 s, and $x = 0.250\cos(4.00t)$ metres, never exceeding 0.250 m.
Show solutionThe two kinematic quantities
$$v = \omega r = (4.00)(0.250) = 1.00\ \mathrm{m/s}$$
steady circular motion, so the speed is the angular rate times the radius and it never changes
the angle grows linearly with time because the rate is constant
$$|x|_{\max} = r = 0.250\ \mathrm{m}$$
a cosine never exceeds one, and it reaches one when the marker is on the axis
Answer $$\boxed{\;v = 1.00\ \mathrm{m/s},\quad T = 1.57\ \mathrm{s},\quad x = 0.250\cos(4.00t)\ \mathrm{m}\;}$$
Check
Check on the period without the formula: at 1.00 m/s the marker covers the circumference $2\pi(0.250) = 1.57$ m in 1.57 s, which is the same number by a different route.
Look at part (b): a quantity going back and forth between $+0.250$ m and $-0.250$ m as a cosine of time, from a body doing nothing but turning steadily. That coincidence is the subject of one of the blocks below.
Notation
symbol
reads as
means
watch out
$x$
ex
the displacement from the equilibrium position, in metres, positive in the direction of stretch
Not the length of the spring and not the stretch from its natural length once a hanging mass is involved.
$A$
ay
the amplitude, the largest value $x$ reaches, in metres
$A$ is positive by definition. It is set by how the motion was started and it never appears in the period.
$T$
tee
the period, the time for one complete there and back trip, in seconds
The same letter meant a tension in the force sections; here a tension is written $F_{T}$.
$f$
eff
the frequency, the number of complete cycles per second, in
$f = 1/T$, never $2\pi/T$. The version with the $2\pi$ is the angular frequency.
$\omega$
omega
the angular frequency, $2\pi f = 2\pi/T$, in radians per second
It is what appears inside the cosine, not $f$, and the block is not rotating even though the symbol says angular.
$\varphi$
phi
the phase constant, the value of the angle inside the cosine at $t=0$, in radians
It is fixed by the starting position and the starting velocity together. One of them alone leaves two candidates.
$\omega t + \varphi$
omega tee plus phi
the phase, the angle inside the cosine at the instant $t$, in radians
The phase advances by $2\pi$ every period, which is what makes the motion repeat.
$k$
kay
the of the spring, in newtons per metre; for other systems, the effective constant read off the net force
In a pendulum the role is played by $mg/L$, which is why the mass cancels.
$\omega_{0}$
omega nought
the natural angular frequency, the one the system would have with no damping and no driving
The damped system oscillates at $\omega'$, which is smaller, and a driven one responds at whatever it is driven at.
$b$
bee
the damping constant in a drag force $F = -bv$, in newton seconds per metre
$b$ multiplies a velocity and $k$ a displacement; the units tell them apart.
$I$
eye
the moment of inertia about the pivot the body actually swings from, in $\mathrm{kg\cdot m^{2}}$
About the pivot, not the centre of mass; the parallel axis theorem carries you between them.
$d$
dee
in a physical pendulum, the distance from the pivot to the centre of mass, in metres
Not the length of the body. For a uniform rod hanging from one end it is half the length.
Conventions used here
Where $x$ is measured from, and which way is positive
In every problem here $x$ is the displacement from the equilibrium position, where the net force is zero, and positive $x$ points along the direction of stretch. For a hanging mass that is the position it rests at, not the end of the unstretched spring. For a pendulum the angle $\theta$ is measured from the downward vertical, positive on the side the bob is displaced to.
Choose any other origin and the net force stops looking like $-kx$, which is the single condition that everything in this section is built on.
The sign that makes the difference between an oscillation and a runaway
The minus sign in $F=-kx$ is not decoration. With a plus sign the force pushes the body further out the further out it goes, and the body never comes back. Every formula in this section assumes the minus, and $k$ itself is always a positive number.
A dropped minus sign gives an equation whose solutions grow instead of oscillating, and the mistake is invisible once you start substituting numbers.
Radians, always, inside the cosine
$\omega$ is in radians per second, so $\omega t + \varphi$ is an angle in radians and the calculator must be in radian mode. Degrees appear in this section only when a swing amplitude is described in words, and they are converted before entering any formula.
The commonest single-line disaster in this material is $\cos^{-1}$ returning degrees and the number being divided by $\omega$ as though it were radians, which is wrong by a factor of 57.3.
The same letter $\omega$ in two different jobs
In the rotational sections $\omega$ was the angular velocity of a body that really turns. Here it is the angular frequency of an oscillation, and the block does not turn at all. Same units, rad/s, and the two meanings coincide in one place only: the reference circle, where the oscillation genuinely is the shadow of something turning.
Same symbol, same units, different physical object. Naming the clash once is cheaper than meeting it in an exam.
Where the zero of stored energy sits
The zero of the spring's stored energy is at the equilibrium position, so the store is $\tfrac12 kx^{2}$ with $x$ measured from there. For a vertical spring this choice quietly swallows gravity: with $x$ measured from the hanging equilibrium, the gravitational and elastic stores together are $\tfrac12 kx^{2}$ plus a constant, and the constant never survives a difference.
It is the reason a hanging mass and a mass on a bench obey the same period formula, and it saves you from carrying a $mgy$ term that cancels.
The constants and the number of digits
$g = 9.80\ \mathrm{m/s^{2}}$ everywhere, as in every earlier section, and no block quietly rounds it to 10. Answers are given to three significant figures; intermediate values carry more, and any value that has been rounded on the way is written out so you can see it.
A period is a square root of a ratio, so a two per cent slip in $g$ moves the answer by one per cent and destroys any check you try to run against a laboratory measurement.
14.1The restoring force: why anything repeats at all
A force that grows in proportion to the displacement and points back to one place is the whole reason motion repeats.
The rotational sections ended with bodies that keep turning. This one is about bodies that keep coming back.
Solvable with what we have
The speed of a block launched by a spring squashed 0.150 m: the store $\tfrac12 kx^{2}$ becomes $\tfrac12 mv^{2}$, and for $k=200$ N/m and $m=0.500$ kg that is 3.00 m/s.
How far a block slides on a rough floor before it stops, from the ledger with $f_k d$ in it.
The force the spring pulls with at any stated stretch, straight from $F=-kx$.
Not solvable yet
How long the block takes to travel from the release point to the middle.
Where the block is 0.100 s after release.
How many times a second the block goes back and forth.
Use the second law and then a constant acceleration formula. At the release point $a = kx/m = (200)(0.150)/0.500 = 60.0\ \mathrm{m/s^{2}}$, so from rest over 0.150 m the time should be $t = \sqrt{2x/a} = \sqrt{2(0.150)/60.0} = 0.0707\ \mathrm{s}$.
Why it fails
The trip really takes 0.0785 s, so the estimate is 10.0% short. That 60.0 m/s$^{2}$ is the acceleration at the release point only; the spring is back at its natural length in the middle, where the acceleration is zero. The constant-acceleration formula was fed the largest number in the trip. The shortfall is 10.0% for every spring, every mass and every pull, which hints that something exact is going on underneath.
DefinitionDefinition 14.1: simple harmonic motion
Conditions
$x$ is measured from the equilibrium position, the one place where the net force vanishes
$k$ is a positive constant, so the force grows in strict proportion to the displacement
the minus sign is part of the statement: the force points back towards $x=0$ from either side
no friction and no drag, otherwise the description holds only for one trip at a time
$$\boxed{\;F_{\rm net} = -kx \qquad\Longleftrightarrow\qquad a = -\frac{k}{m}\,x\;}$$
Pull it twice as far out and it pulls back twice as hard, and it always pulls back rather than out.
Why the vertical spring obeys the same statement
Hang a mass on a spring and let it settle. At that resting position the spring is stretched by some $x_{0}$ with $kx_{0}=mg$. Measure $y$ from there, downwards positive: the spring pulls up with $k(x_{0}+y)$ and the weight down with $mg$, so the net force is $mg-k(x_{0}+y) = -ky$, the first two terms having been made equal by construction. Gravity has vanished, and the only trace it leaves is where the middle of the motion sits.
The same block drawn at three displacements. Pulled out to $+A$ the $\textcolor{#cf222e}{\text{spring force}}$ points back towards the middle; squashed to $-A$ it points back the other way; at the middle it is zero. The $\textcolor{#8250df}{\text{weight and the normal force}}$ cancel throughout, so the whole net force is the horizontal one.
Looks like this, but is not
A ball bounced on a hard floor repeats too, at a steady rhythm you can count, and it looks like the same phenomenon.
The force on the ball is not proportional to anything: it is $mg$ for the whole flight and then something enormous for a millisecond of contact. The test is the amplitude. A ball dropped from four times the height takes twice as long between bounces; the block pulled four times as far takes exactly as long. Repetition alone is not simple harmonic motion; repetition whose rhythm ignores the size of the swing is.
How stiff is the spring that a 0.250 kg mass stretches by 4.90 cm?
A 0.250 kg mass is hung on a light spring and allowed to settle. The spring is then 4.90 cm longer than it was when nothing hung on it. Find the force constant of the spring.
Given
$m = 0.250\ \mathrm{kg}$
static stretch $x_{0} = 0.0490\ \mathrm{m}$
$g = 9.80\ \mathrm{m/s^{2}}$
the mass is at rest
Find
the force constant $k$
SolutionUse the fact that it is sitting still
$$F_{\rm net} = 0 \;\Rightarrow\; k x_{0} = mg$$
at rest means the two vertical forces balance, and this is the only equation the situation offers
solving for $k$ rather than for $x_{0}$ because $k$ is the unknown and everything else is measured
$$k = 50.0\ \mathrm{N/m}$$
newtons per metre, which is the only combination of the given units that can come out
Answer $$\boxed{\;k = 50.0\ \mathrm{N/m}\;}$$
Check
Size check by a different route: 50 N/m means it takes 5 N, about the weight of half a litre of water, to stretch it by 10 cm. That is a soft laboratory spring, which is exactly what sags 4.9 cm under a 250 g mass.
A static measurement, with no motion in it at all, has just handed over the number that will set the rhythm of the motion. That is worth remembering: you can measure $k$ by hanging a mass on the spring and reaching for a ruler.
Does the hanging mass swing about the natural length or somewhere else?
The same 0.250 kg mass on the same 50.0 N/m spring is pulled 3.00 cm below its resting position and released. Show that the net force on it is $-ky$ with $y$ measured from the resting position, and find the size and direction of that force at the moment of release.
Independent check on the claim that gravity has gone: work the same numbers out the long way. The total stretch is $0.0490+0.0300 = 0.0790$ m, so the spring pulls up with $(50.0)(0.0790) = 3.95$ N while the weight pulls down with $(0.250)(9.80) = 2.45$ N. The difference is 1.50 N upwards, the same answer, with $g$ used explicitly.
Every vertical spring problem in this section can now be treated as a horizontal one. Find the resting position, measure from there, and forget that gravity is in the room.
Checkpoint
§14.1 — where the force is biggest●○○○○
A block oscillates on a spring on a smooth bench. A classmate says that the block is being pushed hardest at the moment it is moving fastest, on the grounds that the two must go together.
Given
a block on a spring, no friction
the classmate claims the largest force and the largest speed happen at the same place
Find
(a) Decide whether the claim is true, and say in one sentence where each of the two is largest.
Hint 1/4
You are not being asked to compute anything. Ask where the force is zero and where the block has to be momentarily at rest, and see whether those two places can be the same.
Hint 2/4
The force is $F=-kx$, so its size grows with the distance from the middle. A body that is momentarily not moving is at a turning point of its motion.
Hint 3/4
In this system the block turns round at the two ends of the swing and passes the middle at full speed, and the force law is $F=-kx$ with $x$ measured from the middle.
Hint 4/4
The claim is false: the force is largest at the ends where the speed is zero, and it is zero in the middle where the speed is largest.
Show solutionLocate the force
$$|F| = k|x|$$
the size of the force grows in proportion to the distance from the middle
$$|F|_{\max} \text{ at } x = \pm A, \qquad F = 0 \text{ at } x = 0$$
the ends of the swing and the middle respectively
Locate the speed
$$x = \pm A \;\Rightarrow\; v = 0$$
the block reverses there, so it must pass through zero speed
$$x = 0 \;\Rightarrow\; |v| = v_{\max}$$
everything the spring gave it has become motion, since the store is empty at the natural length
Answer $$\boxed{\;\text{false: the maxima are at opposite ends of the motion}\;}$$
Check
Check by an everyday case that needs no algebra: a child on a swing is momentarily still at the top of the arc, where the pull back to the middle is strongest, and is going fastest at the bottom, where nothing is pulling along the direction of travel.
Whenever a question links a force and a speed in this section, expect them to be a quarter of a cycle apart rather than together.
⚠ Measuring the displacement of a hanging mass from the unstretched length
The natural length is the one the spring has when it is lying on the bench, so it feels like the honest zero, and the resting position looks like an accident of the mass that happens to be hanging there.
The sign carries no number, so it is the first thing to be lost when the equation is copied, and every subsequent line still looks tidy.
wrong$$a = +\frac{k}{m}x$$
right$$a = -\frac{k}{m}x$$
The same information as a graph: the $\textcolor{#cf222e}{\text{force}}$ against the displacement is a straight line through the origin with a negative slope, and that slope is $-k$. Step the displacement and watch the force change sign as it crosses the middle.
14.2Simple harmonic motion: the position at every instant
The cosine is the one shape whose second derivative is itself upside down, which is exactly what the force law demands.
The block failed because the acceleration is different at every point. So stop looking for a number and look for a function.
TheoremTheorem 14.2: the motion that $a=-(k/m)x$ produces
Conditions
the net force obeys $F=-kx$ with $x$ from the equilibrium position
$A$ and $\varphi$ are fixed by where the body is and how fast it is going at one chosen instant, not by the spring
$\omega$ is fixed by the spring and the mass alone, and the amplitude does not appear in it
the argument $\omega t+\varphi$ is in radians
$$\boxed{\;x(t) = A\cos(\omega t + \varphi), \qquad \omega = \sqrt{\frac{k}{m}}, \qquad T = \frac{2\pi}{\omega} = 2\pi\sqrt{\frac{m}{k}}, \qquad f = \frac{1}{T}\;}$$
The position is a cosine of time. Its height is whatever you set it to when you started the motion, and its rate is decided by the stiffness and the mass and by nothing else at all. Stiffer spring, faster; heavier block, slower; and the size of the swing never enters.
Substitute the guess and see what it forces
Try $x = A\cos(\omega t+\varphi)$ with $A$, $\omega$ and $\varphi$ unknown. Differentiating twice gives $a = -A\omega^{2}\cos(\omega t+\varphi)$, which is $-\omega^{2}x$. The law demands $a = -(k/m)x$, and the two agree for every $t$ only if $\omega^{2} = k/m$. So the guess works, for exactly one value of $\omega$. That it is the only answer is a counting argument: a law about a second derivative fixes the motion only once you also say where the body is and how fast it is going at one instant, which is two numbers, and the guess carries exactly two, $A$ and $\varphi$.
The $\textcolor{#1f6feb}{\text{position}}$ of the block of the worked example against time: a cosine of amplitude 4.00 cm repeating every 0.314 s. The $\textcolor{#8250df}{\text{period}}$ is the width of one complete pattern and the $\textcolor{#cf222e}{\text{amplitude}}$ is the height of the band it never leaves.
Looks like this, but is not
A graph that wobbles up and down at a perfectly steady rate but whose peaks get smaller each time still crosses zero at evenly spaced instants and still looks like the picture in the box.
That is a real oscillator with drag in it, and it does not satisfy $a=-\omega^{2}x$: at the same displacement it has a different acceleration going out and coming back, because the drag force reverses with the velocity. The theorem describes the case in which every cycle is an exact copy of the last one. The shrinking version has its own block at the end of this section, and its formula has an extra factor in front of the cosine.
The 0.0785 s that the obvious method missed
Return to the block of the opening: 0.500 kg on a spring of force constant 200 N/m, pulled 0.150 m and released from rest on a smooth bench. Find the angular frequency, the period, the frequency, and the time it takes to reach the middle.
Given
$m = 0.500\ \mathrm{kg}$
$k = 200\ \mathrm{N/m}$
$A = 0.150\ \mathrm{m}$, released from rest
smooth bench
Find
the time from release to the middle, and the three rate quantities on the way
Solution
The constant acceleration formulas are not merely inaccurate here, they are inapplicable: they describe a fixed $a$, and there is no instant at which this $a$ is fixed.
The rate quantities come first and need nothing else
a quarter of a cycle, because a cosine falls from its peak to zero in a quarter of its period
Answer $$\boxed{\;\omega = 20.0\ \mathrm{rad/s},\quad T = 0.314\ \mathrm{s},\quad f = 3.18\ \mathrm{Hz},\quad t = 0.0785\ \mathrm{s}\;}$$
Check
Independent check on why the opening estimate was 10% short, valid for any numbers. The naive time is $\sqrt{2A/a_{\max}}$ with $a_{\max} = \omega^{2}A$, that is $\sqrt{2}/\omega$; the true time is $\pi/2\omega$. Their ratio is $(\pi/2)/\sqrt{2} = 1.1107$, with every symbol cancelled, so the naive method is short by the same 10.0% for every spring and every pull.
Three formulas, and the amplitude of 0.150 m was never used once.
The number the opening asked for is out, and the promised 0.0785 s is on the page. Notice what did the work: the amplitude was given and never needed.
Where is the block 0.100 s after release?
The same block: 0.500 kg on a 200 N/m spring, now pulled 4.00 cm and released from rest at $t=0$. Write the position at every instant and find where the block is 0.100 s later.
Given
$\omega = 20.0\ \mathrm{rad/s}$ from the same $k$ and $m$
$A = 0.0400\ \mathrm{m}$
at $t=0$ the block is at $x=+A$ and at rest
Find
$x(t)$, and the value of $x$ at $t = 0.100$ s
SolutionFix the two free constants from the start of the motion
Independent check without the cosine. The period is 0.314 s, so 0.100 s is 0.318 of a cycle: past the quarter mark, where the block crosses the middle, and short of the half mark, where it reaches the far end. The answer therefore had to lie between 0 and $-4.00$ cm, and it does.
Two questions, one formula. Fixing $\varphi$ from the start of the motion is the step that turns a general shape into this particular block's history.
A block that is pushed rather than released: finding $A$ and $\varphi$
The same spring and block, $\omega = 20.0$ rad/s. This time, at the instant the clock starts, the block is 3.00 cm to the right of the middle and moving to the left at 0.500 m/s. Find the amplitude and the phase constant, and write $x(t)$.
Given
$\omega = 20.0\ \mathrm{rad/s}$
$x_{0} = +0.0300\ \mathrm{m}$ at $t=0$
$v_{0} = -0.500\ \mathrm{m/s}$ at $t=0$
Find
the amplitude, the phase constant, and the position at every instant
Solution
Squaring and adding, rather than solving the two equations one after the other, because the identity $\cos^{2}+\sin^{2}=1$ removes the awkward constant for free.
Write down the two starting conditions
$$x_{0} = A\cos\varphi = 0.0300$$
the position equation evaluated at the instant the clock reads zero
$$v_{0} = -A\omega\sin\varphi = -0.500$$
the velocity equation at the same instant; this is the equation that will decide the sign of $\varphi$
Independent check by putting the answer back into both starting conditions, which were not both used in the same line: $A\cos\varphi = (0.0391)(0.7682) = 0.0300$ m and $-A\omega\sin\varphi = -(0.0391)(20.0)(0.6390) = -0.500$ m/s. Both reproduce the data, which they would not do if the sign of $\varphi$ had been taken the other way.
Any question that hands you a position and a velocity at the same instant is this example. The amplitude is $\sqrt{x_{0}^{2}+(v_{0}/\omega)^{2}}$ and it is always at least as big as the position you were given.
Checkpoint
§14.2 — what a heavier block does to the period●●○○○
Two identical springs hang side by side in a laboratory. One carries a 1.0 kg mass and the other a 4.0 kg mass. Both are set going with the same small pull and released.
Given
two identical springs, so the same $k$
masses of 1.0 kg and 4.0 kg
the same initial pull in both cases
Find
(a) Choose the relation between the period of the heavy one and the period of the light one.
Hint 1/4
Only one quantity differs between the two systems. Find the expression the period depends on and see how that quantity enters it.
Hint 2/4
For a mass on a spring $T = 2\pi\sqrt{m/k}$. The mass sits inside a square root, so a factor in the mass becomes the square root of that factor in the period.
Hint 3/4
Here $k$ is the same for both and the masses are 1.0 kg and 4.0 kg, a factor of four apart, with the same initial pull in both cases.
Hint 4/4
The heavy one takes twice as long per cycle.
Show solutionTake the ratio rather than two separate values
Check at a limit: send the mass towards zero and the period should go to zero, since a spring with nothing on it snaps back instantly. The square root does that, and so does the answer.
Ratios are the fastest tool in this section, because $k$ and the $2\pi$ nearly always cancel.
⚠ Putting the frequency in hertz where the angular frequency belongs
Both are called frequency in ordinary speech and both come out of the same calculation, so the factor of $2\pi$ between them has nowhere obvious to live.
14.3Velocity and acceleration: the same curve, shifted
Each derivative multiplies the height by omega and slides the picture a quarter of a cycle to the left.
We have the position at every instant. Differentiating it is free, and it answers most of the remaining questions.
RuleRule 14.3: the three curves and their maxima
Conditions
the same $A$, $\omega$ and $\varphi$ throughout; these are one motion described three ways
$v$ and $a$ are signed quantities, and the sign is the direction along the axis
the last expression, $v$ against $x$, holds at every instant but cannot tell you which way the body is going
$$\boxed{\;\begin{aligned} x &= A\cos(\omega t+\varphi) \\ v &= -A\omega\sin(\omega t+\varphi), & v_{\max} &= A\omega \\ a &= -A\omega^{2}\cos(\omega t+\varphi) = -\omega^{2}x, & a_{\max} &= A\omega^{2} \\ v &= \pm\,\omega\sqrt{A^{2}-x^{2}} & & \end{aligned}\;}$$
Position, velocity and acceleration are the same wave drawn three times. Each differentiation makes it a factor of omega taller and moves it a quarter of a cycle earlier. The speed is largest in the middle where the force is zero, and the acceleration is largest at the ends where the body is not moving at all.
Where the speed at a given position comes from
Take $x = A\cos\theta$ and $v = -A\omega\sin\theta$ with $\theta = \omega t+\varphi$. Then $\cos\theta = x/A$ and $\sin\theta = -v/(A\omega)$, and since $\cos^{2}\theta+\sin^{2}\theta = 1$ we get $x^{2}/A^{2} + v^{2}/(A\omega)^{2} = 1$. Rearranged, $v = \pm\omega\sqrt{A^{2}-x^{2}}$. The time has been eliminated, which is why this form answers a question that never mentions a clock, and the price is the sign: it gives the speed at a position without saying which of the two passes through that position you are watching.
The $\textcolor{#1f6feb}{\text{position}}$, the $\textcolor{#8250df}{\text{velocity}}$ and the $\textcolor{#cf222e}{\text{acceleration}}$ of one oscillation against a shared time axis. At the dashed lines, a quarter and three quarters of the way through, the position and the acceleration are both zero and the speed is at its largest.
Looks like this, but is not
It is tempting to read the three curves as saying that the block speeds up whenever it is moving in the positive direction and slows down when it moves back, the way a falling body speeds up on the way down.
Speeding up is not about the direction of the velocity, it is about whether the acceleration agrees with it. Here $a = -\omega^{2}x$, so the block speeds up whenever it is heading towards the middle, on either side, and slows down whenever it is heading away, on either side. In one complete cycle it therefore speeds up twice and slows down twice, and the direction of travel changes only twice.
Fastest, and most violently pushed, for a 4.00 cm swing
A 0.500 kg block on a 200 N/m spring swings with an amplitude of 4.00 cm. Find its maximum speed and maximum acceleration, say where each one happens, and find its speed at the moment it is 2.00 cm from the middle.
Independent check on $a_{\max}$ straight from the second law, without any of this section's formulas: at full stretch the spring pulls with $kA = (200)(0.0400) = 8.00$ N, and $8.00/0.500 = 16.0\ \mathrm{m/s^{2}}$. Two routes, same number.
The ratio $\sqrt{3}/2$ at half amplitude is worth carrying: the speed falls off very slowly near the middle and then collapses near the ends.
Reading the phase constant off a description in words
A block on a spring with $\omega = 20.0$ rad/s is timed from the instant it passes the middle moving in the negative direction, with a maximum speed of 0.800 m/s. Find the amplitude and the phase constant, and write down $x(t)$ and $v(t)$.
Given
$\omega = 20.0\ \mathrm{rad/s}$
at $t=0$: $x = 0$ and the block is moving in the $-x$ direction
$v_{\max} = 0.800\ \mathrm{m/s}$
Find
$A$, $\varphi$, and the two functions of time
Solution
Solving for $\varphi$ from the velocity rather than the position, because the position equation has two solutions and the velocity equation has one.
Get the amplitude from the maximum speed
$$v_{\max} = A\omega \;\Rightarrow\; A = \frac{0.800}{20.0} = 0.0400\ \mathrm{m}$$
the maximum speed is the only size information given, so it is the only place the amplitude can come from
Two candidate phases, and the velocity decides between them
Independent check at a later instant: a quarter of a period after the start, $t = T/4 = 0.0785$ s, the block should be at the far negative end. The formula gives $x = -(0.0400)\sin(\pi/2) = -0.0400$ m, which is $-A$ exactly.
Whenever a question starts the clock at the middle rather than at the end, expect a phase of $\pm\pi/2$, and let the direction of travel pick the sign.
Checkpoint
§14.3 — acceleration at the middle of the swing●○○○○
A student writes in a laboratory report that the block is accelerating hardest as it whips through the middle of its swing, because that is where it is moving fastest and the motion looks most violent there.
Given
a block oscillating on a spring
the claim is that the acceleration is largest at $x=0$
Find
(a) Decide whether the claim is true and give the value of the acceleration at the middle.
Hint 1/4
Do not think about the speed at all. Ask what the acceleration is proportional to, and then ask what that quantity is worth at the middle.
Hint 2/4
In this motion $a = -\omega^{2}x$, so the acceleration is fixed by the position and by nothing else.
Hint 3/4
At the middle of the swing $x=0$, and the relation available is $a=-\omega^{2}x$ with $\omega$ some positive number.
Hint 4/4
False: the acceleration at the middle is exactly zero, and it is the speed, not the acceleration, that peaks there.
Show solutionUse the relation between acceleration and position
$$a = -\omega^{2}x$$
this holds at every instant of the motion, so it holds at this one
$$x = 0 \;\Rightarrow\; a = 0$$
the spring is at its natural length, so there is no horizontal force to produce an acceleration
Answer $$\boxed{\;\text{false: } a = 0 \text{ at the middle}\;}$$
Check
Check from the graphs rather than the algebra: the velocity curve is at a peak at that instant, and the slope of a curve at its own peak is zero, and the slope of the velocity is the acceleration.
Whenever speed and acceleration seem to be asked about together in this section, expect the answer to be that they are a quarter of a cycle apart.
⚠ Swapping the two maxima
They differ by a single factor of $\omega$ and the two formulas sit next to each other on every formula sheet, so the eye picks the wrong one under pressure.
⚠ Taking the magnitude and losing the direction of the acceleration
The formula $a=-\omega^{2}x$ looks like a size, and the minus sign is easy to read as decoration rather than as the statement that the acceleration always points at the middle.
wrong$$a = +\omega^{2}x \Rightarrow \text{acceleration points away from the middle}$$
right$$a = -\omega^{2}x \Rightarrow \text{acceleration always points at the middle}$$
⚠ Using the position form of the speed and then asking which way the block is going
The expression $v = \pm\omega\sqrt{A^{2}-x^{2}}$ answers so much with so little that it feels like it must contain everything.
wrong$$v = +\omega\sqrt{A^{2}-x^{2}}\ \text{, so the block moves in the } +x \text{ direction}$$
right$$|v| = \omega\sqrt{A^{2}-x^{2}},\ \text{direction from } v=-A\omega\sin(\omega t+\varphi)$$
14.4Energy in an oscillation: one number traded back and forth
The whole motion carries one fixed amount of energy, and the oscillation is that amount changing hands twice a cycle.
So far every answer needed the clock. Most exam questions do not mention one, and the energy sections already built the tool that ignores it.
TheoremTheorem 14.4: the constant total of a harmonic oscillator
Conditions
no friction and no drag, so the total really is constant rather than slowly falling
$x$ measured from the equilibrium position, so that the store is $\tfrac12 kx^{2}$ with no extra term
for a vertical spring this still holds, because measuring from the hanging position absorbs gravity
The kinetic part and the stored part add up to the same number at every instant of the motion. At the ends of the swing all of it is in the spring; in the middle all of it is in the block; everywhere else it is split. The total is set by the amplitude alone, and it grows as the square of it.
Why the total does not move, in one line of trigonometry
Substitute the two functions of time into the sum. The kinetic part is $\tfrac12 m A^{2}\omega^{2}\sin^{2}(\omega t+\varphi)$ and the stored part is $\tfrac12 kA^{2}\cos^{2}(\omega t+\varphi)$. Since $\omega^{2} = k/m$, the first is $\tfrac12 kA^{2}\sin^{2}$, so the sum collapses to $\tfrac12 kA^{2}$, with no time in it. The energy sections proved this generally, by showing the spring force is conservative; this is the same result read off the motion, and it also gives the value of the constant.
The $\textcolor{#1f6feb}{\text{stored energy}}$ is a parabola and the $\textcolor{#cf222e}{\text{total}}$ is a horizontal line at 0.160 J. The gap between them at any displacement is the $\textcolor{#8250df}{\text{kinetic energy}}$, and the two places where the gap closes are the ends of the swing.
Looks like this, but is not
Because the total is $\tfrac12 kA^{2}$ and the maximum speed is $A\omega$, it looks as though doubling the amplitude does the same thing to everything: twice the swing, twice the speed, twice the energy.
Two of those are right and one is not. Double the amplitude and the maximum speed doubles, because $v_{\max}=A\omega$ is linear in $A$, but the energy quadruples, because $\tfrac12 kA^{2}$ is not. The period does not change at all. Three quantities, three different responses to the same change, and the only way to keep them straight is to look at where $A$ sits in each expression.
The energy of a 4.00 cm swing, and the speed it buys
A 0.500 kg block on a 200 N/m spring swings with an amplitude of 4.00 cm on a smooth bench. Find the total energy of the motion and the maximum speed, using energy alone.
Given
$m=0.500\ \mathrm{kg}$, $k=200\ \mathrm{N/m}$
$A = 0.0400\ \mathrm{m}$
smooth bench, so the total is constant
Find
the total energy and the maximum speed
SolutionEvaluate the total at the place where it is easiest
Independent check by the route that has nothing to do with energy: the previous block gave $v_{\max} = A\omega = (0.0400)(20.0) = 0.800$ m/s from the shape of the motion in time. Energy and calculus agree to three digits, and they used different formulas to get there.
Notice which questions this method cannot touch. It produced a speed and it will produce a position, but it can never produce an instant, because the time was thrown away when the clock was left out.
Where is half the energy still in the spring?
For the same oscillator, amplitude 4.00 cm, find the displacement at which the kinetic energy and the stored energy are equal, and the speed there.
$A = 0.0400\ \mathrm{m}$, total $E = 0.160\ \mathrm{J}$
Find
the position at which the two halves of the energy are equal, and the speed there
Solution
Working with the store rather than with the kinetic part, because the store depends only on $x$, which is what the question wants, so no speed has to be carried through the algebra.
Turn the condition into an equation about the store
$$\tfrac12 kx^{2} = \tfrac12 E = \tfrac12\left(\tfrac12 kA^{2}\right)$$
equal halves means the store holds half the total, which is a statement about $x$ alone
$$x^{2} = \tfrac12 A^{2} \;\Rightarrow\; x = \frac{A}{\sqrt{2}} = 0.707A$$
solving for $x$ rather than for the energies, because the position is what was asked for
which is 71% of the maximum speed at 71% of the amplitude, a coincidence of the square root and not a general rule
Answer $$\boxed{\;x = \pm 2.83\ \mathrm{cm},\qquad v = 0.566\ \mathrm{m/s}\;}$$
Check
Independent check with the position form of the speed, which was derived from the motion and not from energy: $v = \omega\sqrt{A^{2}-x^{2}} = 20.0\sqrt{0.00160-0.00080} = 20.0(0.0283) = 0.566$ m/s. Same number, different tool.
The energy is not shared equally at half the amplitude but at $0.707$ of it, and the reason is that the store grows as the square: at half the amplitude only a quarter of the energy is in the spring.
Checkpoint
§14.4 — what doubling the amplitude changes●●○○○
A demonstrator repeats an experiment with a block on a spring, changing one thing only: the block is pulled to twice the previous distance before being released from rest. Everything else, the block and the spring, is unchanged.
Given
the same block and the same spring
the amplitude is doubled
released from rest in both runs
Find
(a) Choose the statement that correctly describes the second run.
Hint 1/4
Write down the three quantities the question is really about and find the expression for each one. Then look only at how $A$ appears in each.
Hint 2/4
The relevant expressions are $T = 2\pi\sqrt{m/k}$, $v_{\max} = A\omega$ and $E = \tfrac12 kA^{2}$: no $A$, one power of $A$, two powers of $A$.
Hint 3/4
In the second run the amplitude is exactly twice what it was, and $m$ and $k$ are untouched.
Hint 4/4
The energy is four times as large, the maximum speed is twice as large, and the period is exactly the same.
Show solutionRead the power of $A$ in each expression
$$T = 2\pi\sqrt{m/k}$$
no $A$ anywhere, so doubling it changes nothing about the clock
Check with numbers rather than symbols: for $k=200$ N/m the 4.00 cm swing carries 0.160 J and the 8.00 cm swing carries $\tfrac12(200)(0.0800)^{2} = 0.640$ J, which is four times, and the maximum speeds are 0.800 and 1.60 m/s.
If an examiner changes exactly one quantity, the whole question is about which power of that quantity sits in each formula.
⚠ Putting centimetres into the energy formula
Amplitudes are quoted in centimetres because that is what they look like in a laboratory, and the number goes straight into the formula without a stop.
Put a point on a circle of radius A and let it go round at a steady rate. Its horizontal coordinate is a cosine of time with that radius as the amplitude and that rate as the angular frequency. The oscillation is not like the rotation; it is the rotation, seen from the side.
What the picture buys you that the algebra does not
The projection gives the position for free and explains the three constants. The amplitude is the radius. The angular frequency is the rate the point goes round, so one turn is one cycle and $T = 2\pi/\omega$ needs no derivation. The phase constant is where the point was when the clock started, which is why it is an angle and why adding $2\pi$ changes nothing. Even the maximum acceleration comes out: the point has a centripetal acceleration $\omega^{2}A$, and the shadow of that vector is largest when the point is on the axis, at the ends of the swing.
A $\textcolor{#8250df}{\text{point going steadily round}}$ a circle of radius $A$, and its $\textcolor{#1f6feb}{\text{shadow}}$ on the horizontal diameter. The shadow sits at $A\cos\theta$, and since $\theta$ grows at a steady rate the shadow performs exactly the motion of a block on a spring.
Looks like this, but is not
Put the point on a wheel that is speeding up rather than turning steadily. Its shadow still runs back and forth between the same two extremes, and it still looks like an oscillation on a graph.
It is not simple harmonic motion, and the test is the spacing of the crossings. A steady turn crosses the middle at equal intervals for ever; a wheel that is speeding up crosses sooner each time, so there is no period at all. Everything in this section rests on the angle growing linearly with time, which is the same demand as $F=-kx$ read through the picture.
A spot of paint on a wheel, watched edge on
A wheel of radius 0.300 m turns steadily at 45.0 revolutions per minute. A spot of paint on its rim is watched from a long way off in the plane of the wheel, so that only its horizontal position can be seen. Find the amplitude, the period, the angular frequency, the maximum speed and the maximum acceleration of the motion that is seen.
Given
radius $= 0.300\ \mathrm{m}$
45.0 revolutions per minute, constant
the spot is viewed in the plane of the wheel
Find
the five quantities describing the apparent oscillation
SolutionConvert the rate into the two frequencies
$$f = \frac{45.0}{60.0} = 0.750\ \mathrm{Hz}$$
revolutions per minute into revolutions per second, since one revolution is one complete cycle of the shadow
$$T = \frac{1}{f} = 1.33\ \mathrm{s}$$
the shadow completes one there and back trip in exactly one revolution of the wheel
$$\omega = 2\pi f = 4.71\ \mathrm{rad/s}$$
the angular velocity of the wheel and the angular frequency of the shadow are the same number
Read the amplitude off the geometry and take two derivatives
$$A = 0.300\ \mathrm{m}$$
the shadow reaches the axis when the spot does, so the radius is the amplitude
Independent check by a route with nothing to do with oscillation: the real spot is in uniform circular motion, so its acceleration is $\omega^{2}r = (4.71)^{2}(0.300) = 6.66\ \mathrm{m/s^{2}}$ at every instant. At the sides of the wheel that vector lies along the line of sight, so it is the largest acceleration the shadow can show. The two must agree, and they do.
Anything you already know about steady circular motion can be projected and reused here, which is why this picture is worth ten minutes even though nothing depends on it.
Reading the amplitude and the phase off the circle instead of the algebra
A block oscillates with $\omega = 20.0$ rad/s. At the instant the clock starts it is 3.00 cm from the middle on the positive side and moving in the negative direction at 0.500 m/s. Use the reference circle to find the amplitude and the phase constant, without solving simultaneous equations.
Given
$\omega = 20.0\ \mathrm{rad/s}$
$x_{0} = 0.0300\ \mathrm{m}$
$v_{0} = -0.500\ \mathrm{m/s}$
Find
the amplitude and the phase constant, from the geometry
Solution
The picture is chosen here because the sign question, which of two phases is right, is answered by looking at which half of the circle the point is in, rather than by checking a sine.
Place the point on the circle
$$x\text{-coordinate of } P = x_{0} = 0.0300\ \mathrm{m}$$
the shadow is at 3.00 cm, so the point sits somewhere on the vertical line through that value
$$y\text{-coordinate of } P = -\frac{v_{0}}{\omega} = +0.0250\ \mathrm{m}$$
the shadow's speed is the vertical position of the point times $\omega$, so dividing the given speed by $\omega$ locates it; the minus sign puts it in the upper half, where a point going anticlockwise is moving leftwards
The radius and the angle are now a right angled triangle
Independent check against the algebraic route earlier in the section, which solved two simultaneous equations for the same physical situation and got 3.91 cm and 0.695 rad. Two completely different methods, one geometric and one algebraic, landing on the same pair.
If you find the simultaneous equations slippery, draw the circle instead: the amplitude becomes a hypotenuse and the phase becomes an angle you can see.
Checkpoint
§14.5 — what the phase constant is on the circle●●○○○
A student is looking at the reference circle picture and asks what the phase constant actually corresponds to in it, since on the graph of position against time it is just a horizontal shift.
Given
a point going steadily round a circle of radius $A$
the oscillation is the horizontal shadow, $x = A\cos(\omega t+\varphi)$
Find
(a) Choose what $\varphi$ is in the circle picture.
Hint 1/4
Ask what the whole expression $\omega t + \varphi$ is in the picture, and then set the clock to zero and see what is left of it.
Hint 2/4
In the picture $\omega t + \varphi$ is the angle of the point measured round from the positive horizontal axis.
Hint 3/4
Here the clock starts at $t=0$, at which instant the angle of the point is $\omega(0)+\varphi$.
Hint 4/4
It is the angle at which the point was sitting when the clock was started.
Show solutionSet the clock to zero
$$\theta(t) = \omega t + \varphi$$
the angle of the point, growing steadily because the turning is steady
$$\theta(0) = \varphi$$
so the constant is simply where the point was when the timing began
which is why the phase constant is only ever meaningful up to a whole turn
Answer $$\boxed{\;\varphi = \text{the angle of the point at } t=0\;}$$
Check
Check on a case you can see: start the timing at the far right of the circle and the shadow starts at $x=+A$, which is the released from rest case with $\varphi=0$. Start it at the top and the shadow starts at $x=0$ moving left, which was the $\varphi=+\pi/2$ case worked out earlier.
A phase constant is never a physical property of the spring. It is a property of when somebody chose to press the stopwatch.
⚠ Believing the block itself is going round something
The formula is full of angles and the symbol is called angular frequency, so the mind supplies a rotation that is not there.
wrong$$\text{block on a spring: } v = \omega r \text{ with } r \text{ the amplitude, at all times}$$
right$$\text{block on a spring: } |v| = A\omega \text{ only at } x=0,\ \text{and } v=0 \text{ at } x=\pm A$$
⚠ Taking the phase constant to be in degrees because it was found with an inverse tangent
Calculators hand back inverse trigonometric results in whatever mode they are in, and a phase of 39.8 looks as reasonable as a phase of 0.695.
wrong$$x = A\cos(20.0\,t + 39.8)$$
right$$x = A\cos(20.0\,t + 0.695)$$
14.6The pendulum: gravity making a spring out of a string
A swinging bob is a harmonic oscillator whose stiffness gravity supplies, which is why the mass cancels.
Everything so far needed a spring. Here is a system with no spring in it that obeys the same three formulas.
TheoremTheorem 14.6: the periods of the simple and the physical pendulum
Conditions
the swing amplitude is small enough that $\sin\theta$ may be replaced by $\theta$ in radians; the table below prices this
for the simple pendulum the string is light and does not stretch and the bob is small compared with $L$
for the physical pendulum $I$ is taken about the pivot, not about the centre of mass, and $d$ is the pivot to centre distance
For small swings a pendulum keeps time by its length and by gravity alone. Nothing about the bob matters, neither how heavy it is nor what it is made of. For a rigid body the same statement holds with the moment of inertia about the pivot in place of the mass and the distance to the centre of mass in place of the length.
Where the small angle approximation does its work
Only the component of the weight along the arc can change the speed of the bob; the component along the string is cancelled by the tension. That tangential force is $-mg\sin\theta$, the minus sign because it points back towards the vertical. Along the arc the displacement is $s = L\theta$, so for small angles the force is $-(mg/L)s$. That is $F = -ks$ with $k_{\rm eff} = mg/L$, and putting it into $T = 2\pi\sqrt{m/k}$ cancels the mass and leaves $2\pi\sqrt{L/g}$. For a rigid body run the same argument with torques: $\tau = -mgd\sin\theta \approx -mgd\theta$ and $\tau = I\alpha$ give $\omega^{2} = mgd/I$.
The forces on the bob. The $\textcolor{#1f6feb}{\text{tension}}$ and the $\textcolor{#8250df}{\text{component of the weight along the string}}$ cancel, so the entire net force is the $\textcolor{#8250df}{\text{component along the arc}}$, of size $mg\sin\theta$ and always directed back towards the vertical. The $\textcolor{#cf222e}{\text{weight}}$ itself is the only force that has been resolved.
Looks like this, but is not
A conical pendulum, a bob whirled round in a horizontal circle on the end of a string, also has a string of length $L$, also has gravity acting on it, and also has a characteristic time. It looks like the same system.
It is not oscillating at all: nothing about it goes back and forth, the bob travels at constant speed and the string sweeps a cone. Its time for one revolution is $2\pi\sqrt{L\cos\theta/g}$, which depends on the angle, so it fails the amplitude test that defines this whole section. Two systems can share every piece of hardware and still be different problems; what identifies a harmonic oscillator is the shape of the restoring force, never the apparatus.
swing amplitude
amplitude in radians
sine of it
the approximation is low by
the period formula is low by
5 degrees
0.0873
0.0872
0.13%
0.05%
10 degrees
0.1745
0.1736
0.51%
0.19%
15 degrees
0.2618
0.2588
1.14%
0.43%
30 degrees
0.5236
0.5000
4.51%
1.71%
The last column is much smaller than the one before it, and that is the useful part. Replacing the sine by the angle is a 4.5% lie at 30 degrees, but the period it produces is only 1.7% short, because the error enters the period through a square root and through an average over the whole swing rather than at the extreme. A laboratory pendulum released at 10 degrees keeps the textbook period to two parts in a thousand, which is better than most stopwatch work.
How long is a pendulum that ticks once a second?
A pendulum clock ticks once every time the bob passes the middle, so there are two ticks in each complete swing. Find the length of a simple pendulum that ticks once per second, and state its period.
Given
one tick per second, with two ticks in each complete cycle
$g = 9.80\ \mathrm{m/s^{2}}$
small swings, light string
Find
the length of the pendulum
SolutionTurn the ticking rate into a period
$$T = 2\times 1.00 = 2.00\ \mathrm{s}$$
a complete cycle is out and back, which contains two passes through the middle and therefore two ticks
Invert the period formula
$$T = 2\pi\sqrt{\frac{L}{g}} \;\Rightarrow\; L = \frac{gT^{2}}{4\pi^{2}}$$
solving for $L$ rather than substituting blindly, because the length is the unknown
just under a metre, which is why old clock cases are the height they are
Answer $$\boxed{\;L = 0.993\ \mathrm{m},\qquad T = 2.00\ \mathrm{s}\;}$$
Check
Independent check by substituting back into the formula that was inverted: $2\pi\sqrt{0.993/9.80} = 2\pi(0.3183) = 2.00$ s, which recovers the given period and would have caught a factor of $2\pi$ in the wrong place.
A pendulum is a length measured in seconds. If a clock like this runs slow, shorten the pendulum by twice the fractional timing error: 1% slow needs about 2% off the length, because the period follows the square root.
A metre rule swinging from one end
A uniform rule of length 1.00 m is pivoted at one end and allowed to swing through a small angle in a vertical plane. Find its period, and compare it with a simple pendulum of the same length.
Given
uniform rod, $L = 1.00\ \mathrm{m}$, pivoted at one end
$I = \tfrac13 ML^{2}$ about the end, $I_{\rm cm} = \tfrac{1}{12}ML^{2}$
$g = 9.80\ \mathrm{m/s^{2}}$, small swings
Find
the period of the rule, and how it compares with a simple pendulum of the same length
Solution
The physical pendulum formula rather than the simple one, because the mass of a rule is spread along its whole length and there is no single distance at which it can be pretended to sit.
Collect the three quantities the formula needs
$$I = \tfrac13 ML^{2}$$
about the pivot, which is the end; using the value about the centre here is the standard way to get this wrong
$$d = \tfrac12 L$$
the centre of mass of a uniform rod is at its middle, so the pivot to centre distance is half the length
$$m = M$$
the whole mass of the rule, which is about to cancel
Substitute and simplify before reaching for the calculator
shorter than the two seconds a simple pendulum of this length would take
Answer $$\boxed{\;T = 1.64\ \mathrm{s}\;}$$
Check
Independent check through the idea of an equivalent length: the expression $2\pi\sqrt{2L/3g}$ is the simple pendulum formula with an effective length $2L/3 = 0.667$ m, and $2\pi\sqrt{0.667/9.80} = 1.64$ s. The comparison also passes a physical test: a simple pendulum of 1.00 m takes 2.01 s, and the rule is quicker because much of its mass sits well above the far end, closer to the pivot.
Two moments of inertia to keep straight, and only one of them was the right one.
Every physical pendulum can be quoted as an equivalent simple pendulum of length $I/md$. That single number is the most useful thing to compute, because it lets you compare bodies of completely different shapes.
Checkpoint
§14.6 — does a heavier bob swing faster●○○○○
Two pendulums hang from the same beam on strings of the same length. One has a small lead bob and the other has a small wooden bob of a quarter of the mass. Both are set swinging through the same small angle at the same moment.
Given
the same string length for both
the lead bob has four times the mass of the wooden one
the same small swing amplitude, released together
Find
(a) Choose what happens as the two swing.
Hint 1/4
Write down the expression for the period of a simple pendulum and look for the mass in it. Then decide whether its absence is an accident or a consequence.
Hint 2/4
The period is $T = 2\pi\sqrt{L/g}$, in which no mass appears; it dropped out because gravity's pull and the resistance to being accelerated both grow with the mass.
Hint 3/4
Here the lengths are equal, the amplitudes are equal, and only the masses differ, by a factor of four.
Hint 4/4
They stay in step: the periods are identical.
Show solutionFollow the mass through the derivation
$$k_{\rm eff} = \frac{mg}{L}$$
the stiffness that gravity supplies is itself proportional to the mass
the mass in the numerator and the mass hidden in the stiffness cancel exactly, which is the same cancellation that makes all bodies fall at the same rate
Check against a case with a known answer: two objects dropped side by side land together, and this is the same cancellation seen through a string. If the mass did not cancel here it would not cancel there either.
Whenever gravity provides the restoring force, expect the mass to cancel. Whenever a spring provides it, expect the mass to survive.
⚠ Feeding degrees into the small angle step
The amplitude of a swing is naturally described in degrees, and the approximation $\sin\theta\approx\theta$ is then applied to the number that is written down.
With a drag force in it, the oscillation keeps its rhythm almost exactly and loses its height exponentially. If something pushes it at a chosen rate, the response is small at the wrong rates and large near the oscillator's own rate, and how large depends entirely on how much damping there is.
Why the damped shape is stated rather than derived
Checking this solution needs the derivative of a product of an exponential and a cosine, and producing it needs the standard trial solution for a linear equation with constant coefficients. Neither is used elsewhere in this course, and the two facts that matter can be read off the formula without them: the height decays with a time constant $2m/b$, and the frequency is pulled down from $\omega_{0}$ by an amount second order in the damping, negligible whenever the oscillation survives a few cycles.
The $\textcolor{#1f6feb}{\text{dying oscillation}}$ of the worked example, with its $\textcolor{#8250df}{\text{envelope}}$ drawn dashed. The height has halved by the $\textcolor{#cf222e}{\text{marked instant}}$, five and a half cycles in, while the crossings of the middle are still evenly spaced.
Looks like this, but is not
Since damping removes energy, it is natural to expect a damped oscillator to slow down as well, the way a car does when it loses energy to its brakes.
It hardly slows at all. For the oscillator in the example below the frequency falls from 20.000 to 19.996 rad/s, which is two parts in ten thousand, while the height falls by half in five and a half cycles. Damping takes height, not rhythm, and the reason is in the formula: the loss of height is first order in $b$ and the change in frequency is second order. A pendulum clock in a slightly draughty room still keeps time; it just needs winding.
How many cycles before the swing is half as high?
A 0.500 kg block on a 200 N/m spring moves through a liquid that drags on it with $F=-bv$, where $b = 0.400\ \mathrm{N\cdot s/m}$. Find the frequency of the damped motion, the time for the amplitude to fall to half, and how many cycles that is.
Given
$m = 0.500\ \mathrm{kg}$, $k = 200\ \mathrm{N/m}$
$b = 0.400\ \mathrm{N\cdot s/m}$
the undamped value is $\omega_{0} = 20.0\ \mathrm{rad/s}$
Find
the damped angular frequency, the half life of the amplitude, and the number of cycles in it
SolutionThe decay rate first, because everything else is compared with it
Independent check on the approximation made in the last line: the calculation used $T$ from $\omega'$ rather than $\omega_{0}$, and the two differ by 0.02%, which moves 5.51 by about 0.001 cycles. Using the wrong one of the two frequencies is the only way this calculation can go quietly wrong, and here it cannot.
A useful rough rule falls out: the number of cycles before the height halves is about $\tfrac{\ln 2}{\pi}\,m\omega_{0}/b \approx 0.22\,m\omega_{0}/b$, so a light drag buys many cycles and the frequency stays where it was.
At what spin rate does a washing machine shake worst?
The drum of a washing machine, together with its load, has a mass of 40.0 kg and is carried on springs whose combined stiffness is $1.60\times10^{4}\ \mathrm{N/m}$. An out of balance load pushes on the drum once per revolution. Find the spin rate, in revolutions per minute, at which the machine shakes most violently.
Given
$m = 40.0\ \mathrm{kg}$
$k = 1.60\times10^{4}\ \mathrm{N/m}$
the out of balance load drives the system once per revolution
Find
the spin rate at which the response is largest
Solution
The is computed from the spring and the mass and then matched to the driving rate, rather than the other way round, because the driving rate is the quantity being chosen and the natural rate is fixed by the hardware.
cycles per second, which is the form needed before converting to a spin rate
Match the driving rate to it
$$\text{one push per revolution} \;\Rightarrow\; f_{\rm drive} = \text{revolutions per second}$$
so the drum is driven at exactly its own rate when it turns 3.18 times a second
$$(3.18)(60) = 191\ \mathrm{rpm}$$
the response peaks there, and the amplitude at the peak is limited only by the damping in the suspension
Answer $$\boxed{\;\text{worst at about } 191\ \mathrm{rpm}\;}$$
Check
Plausibility check against the machine in the room: domestic machines spin up to somewhere between 800 and 1400 rpm, and the violent shaking everyone has heard happens briefly on the way up, at a couple of hundred revolutions per minute, and then stops. The calculation puts the peak at 191 rpm, which is where that noise is.
Resonance is a matching problem, never a size problem: a small push at the right rate beats a large push at the wrong one. That is why a shaking machine passes through the bad rate quickly rather than pushing less hard.
Checkpoint
§14.7 — what damping mainly costs●●○○○
A laboratory oscillator is run twice with the same spring and the same mass. In the second run it is immersed in a light oil, and it is observed to die away after about twenty swings instead of continuing for several minutes.
Given
the same spring and the same mass in both runs
the second run is damped and dies away in about twenty swings
the damping is light, in the sense that the motion still oscillates
Find
(a) Choose what the oil has mainly changed.
Hint 1/4
Two quantities describe the motion: how high each swing is and how long each swing takes. Ask which of the two the formula changes at first order in the damping.
Hint 2/4
The damped motion is $A_{0}e^{-bt/2m}\cos(\omega' t + \varphi)$ with $\omega' = \sqrt{\omega_{0}^{2}-(b/2m)^{2}}$: the damping enters the height directly and enters the frequency squared.
Hint 3/4
Here the damping is light enough that the motion still oscillates for about twenty swings before it becomes too small to see.
Hint 4/4
The height of each swing collapses while the time per swing is almost exactly what it was.
Show solutionCompare where $b$ enters
$$\text{height} \propto e^{-bt/2m}$$
the damping appears directly in the exponent, so any $b$ at all eats the amplitude
a change of two parts in ten thousand in the rhythm
$$e^{-(0.400)(1.73)} = 0.500$$
against a halving of the height in the same few seconds
Answer $$\boxed{\;\text{the height, not the rhythm}\;}$$
Check
Check by a limit: push the damping up until $b = 2\sqrt{mk}$ and the square root reaches zero, at which point the motion stops oscillating altogether. So the frequency does eventually respond, but only when the damping is strong enough to end the oscillation, not in the light regime described here.
Whenever a question says lightly damped, you are allowed to use the undamped period and spend your attention on the envelope.
⚠ Treating the damping constant as a stiffness
Both are constants of the apparatus written as a single letter in front of a variable, and both make the force bigger.
⚠ Expecting the amplitude at resonance to be infinite
The undamped expression has a zero in the denominator at the natural frequency, and it is tempting to stop reading there.
wrong$$\omega_{\rm drive} \to \omega_{0} \;\Rightarrow\; A \to \infty$$
right$$\omega_{\rm drive} \to \omega_{0} \;\Rightarrow\; A \to \frac{F_{0}}{b\,\omega_{0}}$$
How large the steady response is when the system is driven at a chosen rate, for three amounts of damping. Slide the damping and watch the peak drop and widen; the position of the peak barely moves.
Deciding whether a system is a harmonic oscillator, and finding its stiffness
The apparatus is not a plain spring: a hanging mass, a floating block, a swinging rod, a body in a tunnel. You need to know whether the formulas of this section apply at all before you use any of them.
Find the equilibrium position.
Put the system where the net force, or the net torque, is zero and leave it there. Everything is measured from this place, and nothing is measured from anywhere else.
Displace it by a general amount and write the net force.
Not by a number, by a symbol: displace by $x$, or turn by $\theta$, and write down the whole net force or torque with that symbol in it. Keep the sign of every term.
Look at the shape of what you wrote.
If it is $-(\text{a positive constant})\times x$, the system is a harmonic oscillator and that constant is $k_{\rm eff}$. If the displacement appears squared, or under a sine that you are not allowed to approximate, it is not, and none of this section applies.
Read off the frequency.
For motion along a line, $\omega = \sqrt{k_{\rm eff}/m}$. For motion in an angle, $\omega = \sqrt{\kappa_{\rm eff}/I}$ with the torque written as $-\kappa_{\rm eff}\theta$ and $I$ taken about the pivot.
State the approximation you used and its size.
If you replaced a sine by an angle, say so and say how big the swing may be. This is what makes the answer honest rather than merely correct.
Where it goes wrong
Measuring from the natural length of a spring rather than from the hanging equilibrium, so that a stray $mg$ survives and the force no longer has the right shape.
Using the moment of inertia about the centre of mass in a problem where the body swings about a pivot.
Approximating a sine without checking that the question really does say small swings.
Reading $k_{\rm eff}$ off a force that came out proportional to $x^{2}$, which is not an oscillator of this kind at all.
Getting the amplitude and the phase constant out of a description in words
The question tells you something about the instant $t=0$ and then asks where the body is, or how fast it is going, at some other instant.
Get $\omega$ first, from the hardware only.
$\omega = \sqrt{k/m}$ or $\sqrt{g/L}$ or $\sqrt{mgd/I}$. This never depends on how the motion was started, so it can always be done before you read the rest of the question.
Write the two starting facts as two equations.
$x_{0} = A\cos\varphi$ and $v_{0} = -A\omega\sin\varphi$. Every description in words is one of these two, or both.
Square and add for the amplitude.
$A = \sqrt{x_{0}^{2}+(v_{0}/\omega)^{2}}$. The phase disappears by itself, which is why this is the first thing to compute.
Divide for the phase, then check its sign against the velocity.
$\tan\varphi = -v_{0}/(\omega x_{0})$ gives two candidates a half turn apart. The one you want is the one that makes $v_{0} = -A\omega\sin\varphi$ come out with the sign the question stated.
Use the four standard starts as a check.
Released from rest at $+A$: $\varphi = 0$. Released from rest at $-A$: $\varphi = \pi$. Passing the middle in the positive direction: $\varphi = -\pi/2$. Passing the middle in the negative direction: $\varphi = +\pi/2$.
Where it goes wrong
Getting the phase from the position alone and choosing the wrong one of the two candidates.
Leaving the calculator in degree mode, so that a phase of 0.695 rad is recorded as 39.8 and used inside the cosine.
Taking the amplitude to be the position at $t=0$ when the body was also moving at that instant.
Fixing $\varphi$ from a general instant rather than from the one the clock reads zero at.
Choosing between the energy route and the time route
Every question in this section, before the first line is written. Choosing wrong costs twenty minutes rather than a wrong answer.
Read the question for a clock.
Look for the words instant, after, how long, per second, frequency, phase, number of cycles. If any of them is there, the answer needs $x(t)$ or $v(t)$ and no amount of energy will produce it.
If there is no clock, use energy.
A question that links a speed to a position, or an amplitude to a speed, is answered by $\tfrac12 kA^{2} = \tfrac12 mv^{2}+\tfrac12 kx^{2}$, or by the same statement written as $v = \pm\omega\sqrt{A^{2}-x^{2}}$. No phase constant is needed and none should be computed.
If the question has both, do the energy part first.
Energy usually delivers the amplitude, and the time route needs the amplitude before it can start. The reverse order makes you carry an unknown $A$ through a trigonometric equation.
If something arrives from outside, get the amplitude from the arrival.
A block that slides in and sticks, or a bullet that embeds, sets the amplitude through the energy or the momentum it brings. That is a question from an earlier section wearing this section's clothes.
Where it goes wrong
Solving for the phase constant in a question that never mentions a time.
Trying to get a time out of the energy equation, which has had the clock removed from it on purpose.
Forgetting that the position form of the speed cannot say which way the body is going.
Using conservation of energy across a collision, where it does not hold, instead of momentum.
Speed at 2.00 cm from the middle, by energy
A 0.500 kg block on a 200 N/m spring swings with amplitude 4.00 cm. How fast is it moving as it passes the point 2.00 cm from the middle?
Given
$m=0.500$ kg, $k=200$ N/m, $A=0.0400$ m
the point of interest is $x=0.0200$ m
Find
the speed there
SolutionNo clock in the question, so no clock in the solution
the same statement rearranged, with $\omega = \sqrt{k/m} = 20.0$ rad/s
$$v = 0.693\ \mathrm{m/s}$$
a speed, with no information about which way it is going
Answer $$\boxed{\;v = 0.693\ \mathrm{m/s}\;}$$
Check
Check by proportion: at half the amplitude the speed should be $\sqrt{3}/2 = 0.866$ of the maximum, and $0.866\times0.800 = 0.693$ m/s.
When it first reaches 2.00 cm from the middle, by the time route
The same block and spring, amplitude 4.00 cm, released from rest at the far end at $t=0$. At what instant does it first arrive at the point 2.00 cm from the middle?
Given
$\omega = 20.0$ rad/s, $A = 0.0400$ m
released from rest at $x=+A$ at $t=0$
the point of interest is $x=0.0200$ m
Find
the first instant at which the block is there
SolutionThe question names an instant, so the position function is unavoidable
$$x(t) = A\cos(\omega t), \quad \varphi = 0$$
released from rest at the far end, so the cosine starts at its peak
Check against the period: $T = 0.3142$ s, and $T/6 = 0.0524$ s. Reaching half the amplitude always takes a sixth of a period from a release at the end, whatever the numbers are.
The two questions describe the same block at the same point of the same motion, and the energy route cannot produce the 0.0524 s while the time route cannot produce the 0.693 m/s without first doing extra work to find the phase.
How to tell them apart
Scan the question for a clock. Words like instant, after, how long, how often, phase and cycle send you to $x(t)$; a question that only links a speed to a position is an energy question and asking for a phase constant in it is wasted time.
Doubling the mass on a spring
A 0.500 kg block on a 200 N/m spring is replaced by a 1.000 kg block on the same spring. What happens to the period?
Given
$k = 200$ N/m unchanged
mass goes from 0.500 kg to 1.000 kg
Find
the new period
SolutionTake the ratio, since only one quantity changed
Check by the limit that already has a familiar answer: two bobs of different mass released side by side fall together, and this is the same cancellation with a string attached.
The same change, doubling the mass, stretches the spring's period by 41% and leaves the pendulum's period exactly where it was, because in one case the stiffness is a property of the spring and in the other the stiffness is made by gravity out of the mass itself.
How to tell them apart
Ask what supplies the restoring force. If it is a spring or anything else whose strength was fixed before the body arrived, the mass survives in the period. If it is gravity acting on the same body that is oscillating, the mass cancels.
Scaffolding comes off
The common skeleton
Locate the equilibrium position and agree to measure everything from it.
Write the net force for a general displacement and check that it has the form $-k_{\rm eff}x$; read off $k_{\rm eff}$.
Compute $\omega = \sqrt{k_{\rm eff}/m}$, and from it $T$ and $f$, before touching the rest of the question.
Fix $A$ and $\varphi$ from what is known at one instant, or fix $A$ from the energy if no instant is named.
Answer what was asked: the energy route for a speed at a position, the time route for anything with a clock in it.
Check by a second route, a limit, or the size and units of the answer.
1 · fully worked
Fully worked: period and mid swing speed of a block on a spring
A 0.400 kg block on a smooth bench is attached to a spring of force constant 100 N/m. It is pulled 5.00 cm from the equilibrium position and released from rest. Find the period of the motion and the speed of the block as it passes the point 2.50 cm from the middle.
Given
$m = 0.400\ \mathrm{kg}$, $k = 100\ \mathrm{N/m}$
$A = 0.0500\ \mathrm{m}$, released from rest
the point of interest is $x = 0.0250\ \mathrm{m}$
smooth bench
Find
the period, and the speed at half the amplitude
Solution
The energy route, because the question links a speed to a position and never mentions an instant. The time route would need the phase constant, which would then be thrown away.
Equilibrium and the shape of the force
$$F_{\rm net} = -kx \ \text{with } x \text{ from the natural length}$$
horizontal spring, so the equilibrium is the natural length and there is no gravity term to absorb
$$k_{\rm eff} = k = 100\ \mathrm{N/m}$$
nothing has to be effective here; the spring is the whole story
The rate quantities, which need nothing from the way the motion was started
one subtraction under the root, and the amplitude enters here rather than in the period
$$v = 0.685\ \mathrm{m/s}$$
a speed, not a velocity: the block passes this point twice per cycle going opposite ways
Answer $$\boxed{\;T = 0.397\ \mathrm{s},\qquad v = 0.685\ \mathrm{m/s}\;}$$
Check
Two independent checks. First the maximum speed: $A\omega = (0.0500)(15.81) = 0.791$ m/s, and $0.685/0.791 = 0.866 = \sqrt{3}/2$, which is the exact ratio that must appear at half the amplitude whatever the numbers. Second the energy: $E = \tfrac12(100)(0.0500)^{2} = 0.125$ J, of which $\tfrac12(100)(0.0250)^{2} = 0.03125$ J is still in the spring, leaving 0.0938 J of motion and $v = \sqrt{2(0.0938)/0.400} = 0.685$ m/s.
Every question of this type splits the same way: the period comes from the hardware, and the amplitude comes from how the motion was started.
2 · you write the reasoning
Now an easier problem with the reasoning removed. A simple pendulum has a string of length 0.450 m and a small bob, and it swings through a few degrees. Find its period. The steps are given; write the reason for each one in your own words before opening the model answer. The physics is deliberately lighter than in rung 1, because the work here is explaining, not solving.
The restoring force along the arc is $-mg\sin\theta$, and for a small swing this is $-(mg/L)s$ with $s = L\theta$.
reasoning
Only the component of the weight along the direction of travel can change the speed of the bob; the component along the string is exactly cancelled by the tension, so it does no work and appears nowhere. The approximation replaces $\sin\theta$ by $\theta$, which is legal in radians and costs 0.5% at ten degrees.
So the effective stiffness is $k_{\rm eff} = mg/L$.
reasoning
This is the step that identifies the system as a harmonic oscillator: the force has come out proportional to the displacement along the arc and pointing back towards the middle, which is the definition, and the constant of proportionality is what the definition calls $k$.
Putting that into $\omega = \sqrt{k_{\rm eff}/m}$ gives $\omega = \sqrt{g/L}$.
reasoning
The mass cancels here, and it is worth saying why rather than just watching it go: the stiffness gravity supplies is itself proportional to the mass, so the same $m$ appears above and below the line. This is the pendulum's version of all bodies falling at the same rate.
So $T = 2\pi\sqrt{L/g} = 2\pi\sqrt{0.450/9.80} = 1.35$ s.
reasoning
The period is $2\pi$ divided by the angular frequency, always, and the arithmetic is a square root of 0.0459 which is 0.214, times $2\pi$. The answer is close to one and a third seconds, which is a swing you can count against a clock, and that is the sanity check.
3 · find the buried error
Harder than rung 2, and this worked solution contains exactly two errors. A 0.250 kg block on a spring of force constant 90.0 N/m oscillates with an amplitude of 6.00 cm on a smooth bench. It is released from rest at $x=+A$ at $t=0$. Find (a) the maximum acceleration and (b) the first instant at which the block is at $x = -3.00$ cm. Read the four steps and find the two wrong ones.
Step 1. The angular frequency is $\omega = \sqrt{k/m} = \sqrt{90.0/0.250} = \sqrt{360} = 18.97$ rad/s.
Step 2. The maximum acceleration is $a_{\max} = \omega A = (18.97)(0.0600) = 1.14\ \mathrm{m/s^{2}}$, reached at the ends of the swing.
Step 3. Released from rest at the far end means $\varphi = 0$, so $-0.0300 = 0.0600\cos(\omega t)$, giving $\cos(\omega t) = -0.500$ and $\omega t = 120$.
Step 4. Therefore $t = 120/18.97 = 6.33$ s.
the two buried errors (2)
⚠ step 2
The maximum acceleration is $A\omega^{2}$, not $A\omega$. The expression written down is the maximum speed, 1.14 m/s, and it has been given the units of an acceleration. The correct value is $(0.0600)(360) = 21.6\ \mathrm{m/s^{2}}$.
The two maxima sit next to each other on every formula sheet and differ by a single factor of $\omega$, so under exam pressure the eye takes the first of the pair. Nothing in the line looks wrong, because a number with a decimal point and a unit attached always looks finished.
right
$a_{\max} = A\omega^{2} = (0.0600)(18.97)^{2} = 21.6\ \mathrm{m/s^{2}}$, which is also $kA/m = (90.0)(0.0600)/0.250$, the same number from the second law.
⚠ step 3
The angle 120 is in degrees, but it is about to be divided by $\omega$, which is in radians per second. In radians the phase is $2.094$, so the instant is $2.094/18.97 = 0.110$ s.
The inverse cosine hands back whatever the calculator is set to, and 120 is such a familiar angle that it does not look like a unit error at all. It is the single commonest way to lose a whole question in this material.
right
$\omega t = \cos^{-1}(-0.500) = 2.094$ rad, so $t = 2.094/18.97 = 0.110$ s, which is exactly $T/3$.
4 · the bare problem
§14.2 — amplitude, phase and a first arrival, with no scaffolding●●●●○
The scaffolding is gone. A 0.300 kg block on a smooth bench is attached to a spring of force constant 75.0 N/m. At the instant the clock starts it is 2.00 cm from the equilibrium position on the positive side and moving in the negative direction at 0.400 m/s.
at $t=0$: $x_{0} = +0.0200\ \mathrm{m}$ and $v_{0} = -0.400\ \mathrm{m/s}$
smooth bench
Find
(a) Find the amplitude of the motion.
(b) Find the phase constant.
(c) Find the earliest instant at which the block is at the far end of its swing, $x=-A$.
Hint 1/4
Three quantities are asked for and they come in a forced order: one of them can be computed from the spring alone, and the other two need the two facts about the instant the clock starts.
Hint 2/4
$\omega = \sqrt{k/m}$; then $A = \sqrt{x_{0}^{2}+(v_{0}/\omega)^{2}}$ and $\tan\varphi = -v_{0}/(\omega x_{0})$, with the sign of $\varphi$ decided by $v_{0} = -A\omega\sin\varphi$. The block is at $x=-A$ when the whole phase reaches $\pi$.
Hint 3/4
The numbers again: $m = 0.300$ kg, $k = 75.0$ N/m, $x_{0} = +0.0200$ m and $v_{0} = -0.400$ m/s at the instant the clock reads zero.
Hint 4/4
The amplitude is 3.22 cm, the phase constant is $+0.902$ rad, and the block first reaches the far end at 0.142 s.
Show solution
The amplitude is computed before the phase because squaring and adding removes the phase, while every route to the phase needs the amplitude.
Independent check on (b) by the equation that was not used to get it: $-A\omega\sin\varphi = -(0.0322)(15.81)(0.7845) = -0.400$ m/s, which reproduces the given velocity and would have come out $+0.400$ with the other sign of the phase. Check on (c) by fraction of a period: $0.142/0.397 = 0.357$ of a cycle, between a quarter and a half, exactly where a block starting on the positive side and moving negative should reach the far end.
Written out, this is the whole skeleton: hardware, then amplitude, then phase, then the question. Three of the four steps did not care what was being asked.
Full exam-style question
Exam format: full analysis of a block on a springexam format
A 0.750 kg block on a frictionless horizontal surface is attached to a spring of force constant 120 N/m. The block is pulled 8.00 cm from the equilibrium position and released from rest at $t=0$. (a) Find the period and the frequency. (b) Find the maximum speed and state where it occurs. (c) Find the speed when the block is 4.00 cm from the equilibrium position. (d) Find the time from release until the block first reaches that point.
Given
$m = 0.750\ \mathrm{kg}$
$k = 120\ \mathrm{N/m}$
$A = 0.0800\ \mathrm{m}$, released from rest at $t=0$
frictionless surface
the point of interest in (c) and (d) is $x = 0.0400\ \mathrm{m}$
Find
the period and frequency, the maximum speed, the speed at half amplitude, and the time to first reach it
Solution
Parts (c) and (d) are deliberately the same point of the same motion approached two ways, because that is the choice the examiner is testing: energy when there is no clock, the position function when there is.
radians throughout; in degrees this line would give an answer 57 times too large
Answer $$\boxed{\;T = 0.497\ \mathrm{s},\ f = 2.01\ \mathrm{Hz},\ v_{\max} = 1.01\ \mathrm{m/s},\ v = 0.876\ \mathrm{m/s},\ t = 0.0828\ \mathrm{s}\;}$$
Check
Two independent checks. For (c), by energy: $E = \tfrac12(120)(0.0800)^{2} = 0.384$ J, of which $\tfrac12(120)(0.0400)^{2} = 0.096$ J is still stored, leaving 0.288 J and $v = \sqrt{2(0.288)/0.750} = 0.876$ m/s. For (d): the answer is $T/6 = 0.0828$ s, and a release at the end always reaches half the amplitude after exactly a sixth of a period.
Six formulas, one calculator mode to watch, and the mass was used only once, in the very first line.
If an exam question has four parts on one oscillator, expect the first to be the hardware, the middle two to be energy, and the last to be the one that needs the clock.
Practice
A · concept 4 questions
1§14.1 — a bigger pull and the clock●○○○○
A demonstrator sets a block on a spring going with a 3.00 cm pull, times ten complete cycles, then repeats the whole thing with a 6.00 cm pull. A student predicts, before the second run, that the ten cycles will now take twice as long.
Given
the same block and the same spring in both runs
amplitudes of 3.00 cm and 6.00 cm
the surface is smooth
Find
(a) Decide whether the prediction is right, and give the reason in one sentence.
(b) State one quantity that really does double between the two runs.
Hint 1/4
Write down the expression for the time of one cycle and look for the amplitude in it. Then do the same for the maximum speed.
Hint 2/4
The period is $T = 2\pi\sqrt{m/k}$, which contains no amplitude, while the maximum speed is $v_{\max} = A\omega$, which is proportional to it.
Hint 3/4
Here the amplitude is doubled from 3.00 cm to 6.00 cm and the block and spring are unchanged, so $m$ and $k$ are the same in both runs.
Hint 4/4
The prediction is wrong, the ten cycles take the same time, and the quantity that doubles is the maximum speed.
Show solutionLook for the amplitude in the period
$$T = 2\pi\sqrt{\frac{m}{k}}$$
the expression contains the mass and the stiffness and nothing else
$$\frac{\partial T}{\partial A} = 0$$
there is no $A$ to differentiate, which is the whole content of the answer
Find the quantities that do respond
$$v_{\max} = A\omega \;\Rightarrow\; \times 2$$
linear in the amplitude
$$E = \tfrac12 kA^{2} \;\Rightarrow\; \times 4$$
quadratic in it, so the energy is the most sensitive of the three
Answer $$\boxed{\;\text{the period is unchanged; } v_{\max} \text{ doubles and } E \text{ quadruples}\;}$$
Check
Check by scaling rather than by formula: doubling the amplitude doubles the distance and doubles the acceleration at every corresponding point, and a time built from $\sqrt{\text{distance}/\text{acceleration}}$ is untouched by doubling both.
This property is what makes a pendulum clock possible: it keeps time even as its swing slowly dies away.
2§14.3 — the bob at the end of its swing●●○○○
A pendulum bob is photographed at the exact instant it reaches the far end of its swing and turns round. A student argues that the bob is at rest at that instant, so the net force on it, and therefore its acceleration, must be zero there.
Given
a simple pendulum swinging through a small angle
the instant considered is the far end of the swing, where the bob turns round
Find
(a) Decide whether the acceleration is zero at that instant, and say what it is instead.
Hint 1/4
Separate two different statements: the bob is momentarily not moving, and the bob is not being pushed. Ask whether the second follows from the first.
Hint 2/4
For harmonic motion $a = -\omega^{2}x$, so the acceleration is decided by the position and not by the speed. At the far end the displacement is at its largest.
Hint 3/4
Here the far end of the swing is the point where the displacement has its largest value $A$, and the speed is momentarily zero.
Hint 4/4
False: the acceleration is at its largest there, with magnitude $\omega^{2}A$, pointing back towards the middle.
Show solutionApply the relation between acceleration and position
$$a = -\omega^{2}x$$
the acceleration depends on where the bob is, not on how fast it is going
$$x = A \;\Rightarrow\; |a| = \omega^{2}A$$
the largest value in the whole motion, pointing back towards the middle
Test the alternative claim
$$v = 0 \ \text{and}\ a = 0 \;\Rightarrow\; \text{permanently at rest}$$
if both were zero the bob would stay there for ever, which is not what pendulums do
Answer $$\boxed{\;\text{false: } |a| = \omega^{2}A \text{ at the far end}\;}$$
Check
Check with a component of the weight, using nothing from this section: at an angle $\theta$ the tangential force is $mg\sin\theta$, which is largest at the largest angle, that is at the ends of the swing, and it is zero at the bottom.
Turning round always requires the largest acceleration of the motion. The two ends of an oscillation are where the pushing happens.
3§14.3 — where the speed is half its maximum●●●○○
A block oscillates on a spring with amplitude $A$. A question on a past paper asks for the displacement at which the block is moving at exactly half its maximum speed, and offers four candidate positions.
Given
amplitude $A$, smooth surface
the condition is $\vert v\vert = \tfrac12 v_{\max}$
Find
(a) Choose the displacement at which the speed is half the maximum.
Hint 1/4
You need the one relation that links a speed to a position without any time in it. Write it, then impose the condition and solve for the position.
Hint 2/4
The relation is $v = \pm\omega\sqrt{A^{2}-x^{2}}$, and the maximum speed is $A\omega$, reached at $x=0$.
Hint 3/4
The condition to impose is $\omega\sqrt{A^{2}-x^{2}} = \tfrac12 A\omega$, with $A$ the amplitude of the motion.
Hint 4/4
The speed is half its maximum at $x = \tfrac{\sqrt3}{2}A$, which is $0.866A$.
Show solutionImpose the condition on the position form of the speed
$$\omega\sqrt{A^{2}-x^{2}} = \tfrac12 A\omega$$
the position form is chosen because the answer wanted is a position and no time is mentioned
$$A^{2}-x^{2} = \tfrac14 A^{2}$$
the $\omega$ cancels, so the answer cannot depend on the spring at all
$$x = \pm\frac{\sqrt3}{2}A = \pm 0.866A$$
two positions, one on each side, as the symmetry of the motion requires
Answer $$\boxed{\;x = \pm 0.866\,A\;}$$
Check
Check the complementary case, which is a known result: at $x = 0.5A$ the speed is $\omega A\sqrt{1-0.25} = 0.866\,v_{\max}$. The two numbers 0.5 and 0.866 simply swap places, which they must, because the relation is a circle.
Speed and position in this motion are related the way the two sides of a right angled triangle are, which is why 0.866 and 0.5 keep trading places.
4§14.5 — two motions half a turn out of phase●●●○○
Two identical blocks on identical springs are set going with the same amplitude and the same period, but with phase constants that differ by exactly $\pi$. A student is asked what this means physically and offers four descriptions.
Given
same $A$, same $\omega$
phase constants differing by $\pi$
both motions of the form $A\cos(\omega t+\varphi)$
Find
(a) Choose the description that is correct.
Hint 1/4
Use the circle picture. Two phase constants are two starting angles, so ask where on the circle the two points are relative to each other and stay relative to each other.
Hint 2/4
On the reference circle the phase is the angle of the point, and $\cos(\theta+\pi) = -\cos\theta$ for every angle $\theta$.
Hint 3/4
Here the two angles differ by $\pi$ at $t=0$ and both grow at the same rate $\omega$, so they differ by $\pi$ for ever.
Hint 4/4
The two blocks are always on opposite sides of the middle, at equal distances, so when one is at the far end the other is at the near end.
the identity holds at every instant, so the relation is permanent and not just true at the start
$$\varphi \to \varphi+\pi \iff t \to t + \tfrac{T}{2}$$
half a period advances the phase by $\pi$, so the difference can also be read as a delay of half a cycle
Answer $$\boxed{\;x_{2}(t) = -x_{1}(t) \text{ at every instant}\;}$$
Check
Check on the circle: two points at opposite ends of a diameter stay at opposite ends of a diameter as the circle turns, so their shadows are always equal and opposite. The geometric picture and the trigonometric identity give the same statement.
Phase differences are angles. A difference of $\pi$ is opposite, a difference of $\pi/2$ is a quarter cycle apart, and a difference of $2\pi$ is no difference at all.
B · computation 7 questions
1§14.2 — from a measured period to the stiffness●●○○○
A 0.350 kg mass is hung on a light vertical spring and set oscillating. A student times twenty complete cycles with a stopwatch and gets 12.8 s.
Given
$m = 0.350\ \mathrm{kg}$
twenty complete cycles in 12.8 s
light spring, small oscillations, no damping worth counting
Find
(a) Find the period and the frequency of the motion.
(b) Find the angular frequency.
(c) Find the force constant of the spring.
Hint 1/4
Twenty cycles were timed rather than one, which is a measurement trick rather than a physics one. Undo it first, then work backwards from the period to the spring.
Hint 2/4
$T$ is the time for one cycle, $f = 1/T$ and $\omega = 2\pi/T$; and since $\omega = \sqrt{k/m}$, the stiffness is $k = m\omega^{2}$.
Hint 3/4
The numbers again: $m = 0.350$ kg and twenty complete cycles took 12.8 s.
the version that goes inside a cosine, larger than $f$ by the factor $2\pi$
Invert the frequency relation to reach the spring
$$\omega = \sqrt{\frac{k}{m}} \;\Rightarrow\; k = m\omega^{2}$$
solving for the stiffness, which is the only unmeasured quantity left
$$k = (0.350)(96.38) = 33.7\ \mathrm{N/m}$$
a soft spring: it stretches about 10 cm under a 0.35 kg mass
Answer $$\boxed{\;T = 0.640\ \mathrm{s},\ f = 1.56\ \mathrm{Hz},\ \omega = 9.82\ \mathrm{rad/s},\ k = 33.7\ \mathrm{N/m}\;}$$
Check
Independent check by the static route: a spring of 33.7 N/m carrying 0.350 kg would hang with a stretch of $mg/k = (0.350)(9.80)/33.7 = 0.102$ m, and a 10 cm sag is exactly what a spring that oscillates at one and a half hertz looks like on a laboratory stand.
Timing many cycles and dividing is worth doing in a real measurement, because the human error in starting and stopping the watch is divided by twenty as well.
2§14.1 — a period read off a sag●●●○○
A mass is hung on a light vertical spring and settles 2.50 cm lower than the position of the free end of the spring before the mass was attached. It is then pulled down a little and released.
Given
static sag $x_{0} = 0.0250\ \mathrm{m}$
neither the mass nor the force constant is given
$g = 9.80\ \mathrm{m/s^{2}}$, small oscillations
Find
(a) Find the period of the oscillation.
(b) Explain in one sentence why the mass is not needed.
Hint 1/4
Two unknowns and one measurement looks impossible, so look for a combination of the two unknowns that the measurement gives you directly, and check whether the period needs only that combination.
Hint 2/4
At rest $kx_{0} = mg$, so $m/k = x_{0}/g$; and the period is $T = 2\pi\sqrt{m/k}$, which needs exactly that ratio and nothing else.
Hint 3/4
Here the sag is $x_{0} = 0.0250$ m and $g = 9.80\ \mathrm{m/s^{2}}$.
Hint 4/4
$T = 2\pi\sqrt{x_{0}/g} = 0.317$ s, and the mass is not needed because only the ratio $m/k$ enters.
Show solution
Working with the ratio rather than solving for $m$ and $k$ separately, because the question supplies one equation and two unknowns and only the ratio is determined.
Independent check by inventing numbers that fit and using the ordinary route: take $m = 0.100$ kg, then $k = mg/x_{0} = 39.2$ N/m and $T = 2\pi\sqrt{0.100/39.2} = 0.317$ s. Take $m = 1.00$ kg instead and $k$ becomes 392 N/m, giving the same 0.317 s.
The expression $T = 2\pi\sqrt{x_{0}/g}$ is the pendulum formula with the sag in place of the length, and that is not a coincidence: in both systems the period is set by a length and by gravity.
3§14.3 — speeds and accelerations at a given point●●●○○
A body moves in simple harmonic motion with an amplitude of 0.120 m and a period of 0.500 s.
Given
$A = 0.120\ \mathrm{m}$
$T = 0.500\ \mathrm{s}$
the point of interest is $x = 0.0600\ \mathrm{m}$
Find
(a) Find the angular frequency, the maximum speed and the maximum acceleration.
(b) Find the speed and the acceleration at the instant the body is 0.0600 m from the middle.
Hint 1/4
Every part of this question comes from the same three expressions, so get the angular frequency first and then read the rest off. Nothing here needs a phase constant.
Hint 2/4
$\omega = 2\pi/T$, then $v_{\max} = A\omega$ and $a_{\max} = A\omega^{2}$; at a stated position, $v = \omega\sqrt{A^{2}-x^{2}}$ and $a = -\omega^{2}x$.
Hint 3/4
The numbers again: $A = 0.120$ m, $T = 0.500$ s and the position asked about is $x = 0.0600$ m, which is half the amplitude.
Hint 4/4
$\omega = 12.6$ rad/s, $v_{\max} = 1.51$ m/s, $a_{\max} = 18.9\ \mathrm{m/s^{2}}$, and at half amplitude $v = 1.31$ m/s with $a = -9.47\ \mathrm{m/s^{2}}$.
Show solutionThe angular frequency, then the two maxima
Independent check on the two values at half amplitude by their ratios to the maxima, which must be exactly $\sqrt{3}/2 = 0.866$ for the speed and exactly $1/2$ for the acceleration whatever the numbers: $1.307/1.508 = 0.866$ and $9.47/18.95 = 0.500$.
The acceleration is proportional to the position and the speed is not, which is why halving the position halves one of them and barely touches the other.
4§14.4 — splitting the energy of an oscillation●●●○○
A 0.800 kg block on a smooth bench is attached to a spring of force constant 250 N/m and oscillates with an amplitude of 5.00 cm.
Given
$m = 0.800\ \mathrm{kg}$
$k = 250\ \mathrm{N/m}$
$A = 0.0500\ \mathrm{m}$
smooth bench
Find
(a) Find the total energy of the motion and the maximum speed.
(b) Find the displacement at which the kinetic and stored energies are equal.
(c) Find the speed of the block when it is 2.50 cm from the middle.
Hint 1/4
Every part is the same one equation evaluated somewhere different, so start by writing the total in the place where only one of its two terms survives.
Hint 2/4
$E = \tfrac12 kA^{2}$ at the end of the swing, $E = \tfrac12 mv_{\max}^{2}$ at the middle, and $\tfrac12 kx^{2} = \tfrac12 E$ defines the equal split.
Hint 3/4
The numbers again: $k = 250$ N/m, $A = 0.0500$ m and $m = 0.800$ kg, and part (c) asks about $x = 0.0250$ m.
Hint 4/4
$E = 0.313$ J with $v_{\max} = 0.884$ m/s, the split is equal at $x = 3.54$ cm, and at 2.50 cm the speed is 0.765 m/s.
Show solution(a) The total where the block is at rest, then where the spring is slack
half the amplitude again, so the $\sqrt3/2$ ratio should appear
$$v = 0.765\ \mathrm{m/s}$$
and indeed $0.765/0.884 = 0.866$
Answer $$\boxed{\;E = 0.313\ \mathrm{J},\ v_{\max} = 0.884\ \mathrm{m/s},\ x = 3.54\ \mathrm{cm},\ v = 0.765\ \mathrm{m/s}\;}$$
Check
Independent check on (c) through energy rather than through $\omega$: the store at 2.50 cm is $\tfrac12(250)(0.0250)^{2} = 0.0781$ J, leaving $0.3125-0.0781 = 0.2344$ J of motion, and $v = \sqrt{2(0.2344)/0.800} = 0.765$ m/s.
Notice that (b) came out as a pure fraction of the amplitude with no numbers in it. Whenever an energy question asks where, expect the answer to be a fraction of $A$ rather than a length.
5§14.6 — a pendulum taken somewhere with less gravity●●●○○
A simple pendulum in a laboratory on Earth has a period of 1.20 s. The same pendulum, unchanged, is later used on the Moon, where the free fall acceleration is 1.62 m/s squared.
Given
period on Earth $T_{E} = 1.20\ \mathrm{s}$
$g_{E} = 9.80\ \mathrm{m/s^{2}}$
$g_{M} = 1.62\ \mathrm{m/s^{2}}$
the same pendulum, so the same length, small swings
Find
(a) Find the length of the pendulum.
(b) Find its period on the Moon.
(c) State whether a pendulum clock taken there would run fast or slow.
Hint 1/4
Part (a) is an inversion of the period formula. For part (b), notice that the length is the same in both places, so a ratio will save you from using the length at all.
Hint 2/4
$T = 2\pi\sqrt{L/g}$, so $L = gT^{2}/4\pi^{2}$; and for a fixed length $T \propto 1/\sqrt{g}$, so $T_{M}/T_{E} = \sqrt{g_{E}/g_{M}}$.
Hint 3/4
The numbers again: $T_{E} = 1.20$ s, $g_{E} = 9.80\ \mathrm{m/s^{2}}$ and $g_{M} = 1.62\ \mathrm{m/s^{2}}$.
Hint 4/4
$L = 0.357$ m, $T_{M} = 2.95$ s, and a clock driven by it would run slow.
Show solution
The ratio in part (b) rather than a second substitution, because it removes the length and therefore removes any dependence on the rounding done in part (a).
(a) Invert the formula
$$T = 2\pi\sqrt{\frac{L}{g}} \;\Rightarrow\; L = \frac{gT^{2}}{4\pi^{2}}$$
solving for the length, which is the quantity that stays the same in both places
Independent check on (b) by the direct route: $2\pi\sqrt{0.357/1.62} = 2\pi(0.4695) = 2.95$ s, using the length from part (a) rather than the ratio, and the two agree.
Anything that changes $g$ changes a pendulum clock: altitude, latitude, and in principle the tides. This is why the best mechanical clocks were eventually built around springs and quartz instead.
6§14.6 — a disc swinging from a point on its rim●●●●○
A uniform disc of radius 0.150 m hangs from a small frictionless pivot at a point on its rim, in a vertical plane, and is set swinging through a small angle.
Given
uniform disc, radius $R = 0.150\ \mathrm{m}$
pivoted at a point on the rim
$I_{\rm cm} = \tfrac12 MR^{2}$ about the centre
$g = 9.80\ \mathrm{m/s^{2}}$, small swings
Find
(a) Find the moment of inertia of the disc about the pivot, in terms of $M$ and $R$.
(b) Find the period of the swinging.
(c) Find the length of the simple pendulum that would keep the same time.
Hint 1/4
The body is not a point mass on a string, so the simple pendulum formula does not apply. Two quantities have to be assembled before the right formula can be used: something about how the mass is distributed about the pivot, and how far the centre of mass is from it.
Hint 2/4
$T = 2\pi\sqrt{I/(mgd)}$ with $I$ about the pivot, obtained from $I = I_{\rm cm}+Md^{2}$, and $d$ the pivot to centre distance.
Hint 3/4
The numbers again: $R = 0.150$ m, the pivot is on the rim so $d = R$, and $I_{\rm cm} = \tfrac12 MR^{2}$.
Hint 4/4
$I = \tfrac32 MR^{2}$, $T = 2\pi\sqrt{3R/2g} = 0.952$ s, and the equivalent simple pendulum is 0.225 m long.
Show solution
The physical pendulum formula rather than the simple one, because the mass of a disc is spread over a whole surface and there is no single point at which it can be pretended to sit.
Independent check on (b) using the answer to (c) and the simple pendulum formula, which is a completely different expression: $2\pi\sqrt{0.225/9.80} = 2\pi(0.1515) = 0.952$ s. It also passes a sanity test on size: the equivalent length 0.225 m is larger than the radius but smaller than the diameter, which is where a disc's should sit.
The equivalent length $I/Md$ is the number worth remembering for any swinging body, because it turns every physical pendulum back into a simple one.
7§14.7 — counting the cycles of a dying oscillation●●●●○
A 0.400 kg block on a spring of force constant 90.0 N/m oscillates in a liquid that exerts a drag force $F = -bv$ with $b = 0.240\ \mathrm{N\cdot s/m}$.
Given
$m = 0.400\ \mathrm{kg}$
$k = 90.0\ \mathrm{N/m}$
$b = 0.240\ \mathrm{N\cdot s/m}$
Find
(a) Find the undamped angular frequency and the damped one, and comment on the difference.
(b) Find the time for the amplitude to fall to a quarter of its starting value.
(c) Find how many complete cycles that takes.
Hint 1/4
Two rates control everything here and they should be computed first: the rate at which the block oscillates and the rate at which the envelope decays. Only then does the question about the amplitude become arithmetic.
Hint 2/4
$\omega_{0}=\sqrt{k/m}$, the decay rate is $b/2m$, and $\omega' = \sqrt{\omega_{0}^{2}-(b/2m)^{2}}$. The envelope is $e^{-(b/2m)t}$, so a fall to a fraction $\alpha$ takes $t = \ln(1/\alpha)/(b/2m)$.
Hint 3/4
The numbers again: $m = 0.400$ kg, $k = 90.0$ N/m and $b = 0.240\ \mathrm{N\cdot s/m}$, and the fall asked for in (b) is to one quarter.
Hint 4/4
$\omega_{0}=15.0$ and $\omega'=15.0$ rad/s to three figures, the quarter is reached at 4.62 s, and that is 11.0 complete cycles.
Show solution(a) The oscillation rate and the decay rate
logarithms, because the unknown sits in the exponent
(c) Convert the time into a count
$$T = \frac{2\pi}{15.0} = 0.419\ \mathrm{s}$$
using $\omega'$, which here is indistinguishable from $\omega_{0}$
$$N = \frac{4.62}{0.419} = 11.0\ \text{cycles}$$
eleven complete trips before the swing is a quarter of what it was
Answer $$\boxed{\;\omega' = 15.0\ \mathrm{rad/s},\qquad t = 4.62\ \mathrm{s},\qquad N = 11.0\ \text{cycles}\;}$$
Check
Independent check on (b) and (c) through the half life: the amplitude halves after $\ln2/0.300 = 2.31$ s, and a quarter is two halvings, so 4.62 s, which is exactly twice. In cycles that is 5.51 and then 5.51 again, giving 11.0.
For lightly damped systems the useful quantity is the number of cycles rather than the time, because it does not change when you swap the spring for a stiffer one and the damping in proportion.
C · exam level 4 questions
1§14.2 — a block struck as it sits at rest●●●●○
A 0.250 kg block rests at the equilibrium position on a smooth horizontal surface, attached to a spring of force constant 160 N/m. At $t=0$ it is struck and sent off in the positive direction at 1.20 m/s. Exam format: four parts, and the later ones use the earlier ones.
Given
$m = 0.250\ \mathrm{kg}$
$k = 160\ \mathrm{N/m}$
at $t=0$: $x = 0$ and $v = +1.20\ \mathrm{m/s}$
smooth surface
Find
(a) Find the period and the frequency of the resulting motion.
(b) Find the amplitude.
(c) Find the maximum acceleration and say where it occurs.
(d) Find the position of the block at $t = 0.0500$ s.
Hint 1/4
Parts (a) and (b) come from two different places: one from the hardware alone and one from the way the motion was started. Part (d) needs a phase constant, and the description in words fixes it without any algebra.
Hint 2/4
$\omega = \sqrt{k/m}$, then $A = v_{\max}/\omega$ since the block was struck at the middle, $a_{\max} = A\omega^{2}$, and $x = A\cos(\omega t + \varphi)$ with $\varphi = -\pi/2$ for a start at the middle moving positive, which is the same as $x = A\sin(\omega t)$.
Hint 3/4
The numbers again: $m = 0.250$ kg, $k = 160$ N/m, the block starts at $x=0$ moving at $+1.20$ m/s, and part (d) asks about $t = 0.0500$ s.
Hint 4/4
$T = 0.248$ s and $f = 4.03$ Hz, $A = 4.74$ cm, $a_{\max} = 30.4\ \mathrm{m/s^{2}}$ at the two ends, and $x(0.0500) = +4.52$ cm.
Show solution
The sine form in part (d) rather than a cosine with a phase constant, because a start at the middle is exactly what a sine describes and it removes one place to make a sign error.
Independent check on (b) by energy, which never mentions $\omega$: the block starts with $\tfrac12(0.250)(1.20)^{2} = 0.180$ J, all of it in the spring at the end of the swing, so $\tfrac12(160)A^{2} = 0.180$ and $A = 0.0474$ m. Check on (d): $0.0500/0.2483 = 0.201$ of a cycle, just under the quarter at which the block reaches the far end, so the answer must be positive and just under $A$.
Struck at the middle means the given speed is the maximum speed, which is the shortest possible route to an amplitude. Struck anywhere else and you are back to squaring and adding.
2§14.6 — a pendulum solved twice, by energy and as an oscillator●●●●○
A simple pendulum has a string of length 0.800 m and a small bob. It is pulled aside until the string makes an angle of 12.0 degrees with the vertical and released from rest. Exam format: the last part is the one that carries the marks.
Given
$L = 0.800\ \mathrm{m}$
released from rest at $\theta_{0} = 12.0^{\circ}$
$g = 9.80\ \mathrm{m/s^{2}}$
$\cos 12.0^{\circ} = 0.97815$
Find
(a) Find the period of the swinging.
(b) Find the speed of the bob at the lowest point using conservation of energy.
(c) Find the same speed by treating the motion as simple harmonic, with the amplitude measured along the arc.
(d) Compare the two answers and say which is the approximation and why.
Hint 1/4
Parts (b) and (c) are the same physical quantity reached along two different roads, and the interesting part of the question is that they do not have to agree exactly. Set up both before evaluating either.
Hint 2/4
Energy: the bob rises by $h = L(1-\cos\theta_{0})$, so $v = \sqrt{2gh}$. Oscillator: $\omega = \sqrt{g/L}$ and the amplitude along the arc is $A = L\theta_{0}$ with $\theta_{0}$ in radians, so $v_{\max} = A\omega$.
Hint 3/4
The numbers again: $L = 0.800$ m, $\theta_{0} = 12.0^{\circ} = 0.20944$ rad, $\cos 12.0^{\circ} = 0.97815$ and $g = 9.80\ \mathrm{m/s^{2}}$.
Hint 4/4
$T = 1.80$ s; the energy route gives 0.585 m/s and the oscillator route gives 0.586 m/s, which is 0.18% higher, and the oscillator route is the approximate one.
Show solution
Doing part (b) before part (c), so that the exact answer exists before the approximate one is compared with it.
Independent check that the 0.18% is genuinely the small angle error and not an arithmetic slip: the leading correction to the sine gives an error of about $\theta_{0}^{2}/24$ in this comparison, which for $\theta_{0} = 0.209$ rad is $0.0018$, that is 0.18%. The predicted size and the measured gap agree.
Approximations in physics are not sloppiness, they are a trade with a price attached. Here the price is quoted: two parts in a thousand at twelve degrees.
3§14.7 — a footbridge that a crowd can shake●●●●○
A footbridge is found to bounce with a natural frequency of 1.0 Hz. Engineers are worried because people walking across it take about one step per second, and each step delivers a small push.
Given
the bridge bounces at a natural frequency of 1.0 Hz
walkers deliver about one push per second
the bridge has some damping but not much
Find
(a) Choose the statement that best describes what will happen and what would fix it.
Hint 1/4
Two rates are given and they are equal. Ask what the response of a driven oscillator does when the driving rate matches its natural rate, and then ask which quantity in the formula limits the response there.
Hint 2/4
A driven oscillator responds most strongly when the driving rate matches its natural rate, and at that peak the amplitude is limited only by the damping.
Hint 3/4
Here the natural rate of the bridge and the rate of the pushes are both about one per second, and the damping of the structure is small.
Hint 4/4
The bridge is being driven at resonance, so small pushes build up a large amplitude, and the practical cure is to add damping.
Show solutionCompare the two rates
$$f_{\rm drive} = f_{0} = 1.0\ \mathrm{Hz}$$
the definition of resonance is this equality, not any statement about the size of the push
add dampers, which is what is actually done to bridges
$$\omega_{0} = \sqrt{k/m} \ \text{moved away from } \omega_{\rm drive}$$
stiffen the structure or change its mass so that the peak is no longer where the walkers are
Answer $$\boxed{\;\text{resonance: small pushes at the matching rate, cured by damping}\;}$$
Check
Check by the shape of the response curve: well away from the natural rate the amplitude is close to the deflection a steady push of the same size would produce. Only within a narrow band around the natural rate is the response many times larger, and the width of that band is itself set by the damping.
Resonance questions are always about a comparison of two rates. Compute both, compare them, and only then think about sizes.
4§14.2 — reading an oscillation off a table of measurements●●●○○
A block on a spring is photographed at regular intervals and its displacement from the equilibrium position is recorded. The measurements are listed below, and the spring is known to have a force constant of 40.0 N/m.
Given
$k = 40.0\ \mathrm{N/m}$
$t = 0$ s, $x = +5.00$ cm
$t = 0.10$ s, $x = +3.54$ cm
$t = 0.20$ s, $x = 0.00$ cm
$t = 0.30$ s, $x = -3.54$ cm
$t = 0.40$ s, $x = -5.00$ cm
$t = 0.50$ s, $x = -3.54$ cm
$t = 0.60$ s, $x = 0.00$ cm
$t = 0.70$ s, $x = +3.54$ cm
$t = 0.80$ s, $x = +5.00$ cm
Find
(a) Find the amplitude and the period from the table.
(b) Find the frequency and the angular frequency.
(c) Find the mass of the block.
(d) Write the displacement as a function of time.
Hint 1/4
Do not compute anything at first. Read two numbers straight off the table: the largest displacement that appears, and the time after which the pattern starts repeating exactly.
Hint 2/4
The amplitude is the largest magnitude in the table and the period is the time to return to the same displacement moving the same way. Then $f = 1/T$, $\omega = 2\pi/T$, and from $\omega = \sqrt{k/m}$ the mass is $m = k/\omega^{2}$.
Hint 3/4
The table runs from $t=0$ to $t=0.80$ s in steps of 0.10 s, starting at $+5.00$ cm, and $k = 40.0$ N/m.
Hint 4/4
$A = 5.00$ cm, $T = 0.800$ s, $f = 1.25$ Hz, $\omega = 7.85$ rad/s and $m = 0.648$ kg, so $x = (5.00\ \mathrm{cm})\cos(7.85\,t)$.
Show solution
The period is taken from the return to the same displacement moving the same way, rather than from the gap between the two extremes, because the second of those is half a period and is the standard way to lose this question.
(a) Read, do not compute
$$A = 5.00\ \mathrm{cm}$$
the largest magnitude that appears anywhere in the table, reached at $t=0$, $0.40$ and $0.80$ s
$$T = 0.800\ \mathrm{s}$$
the block is at $+5.00$ cm at $t=0$ and again at $t=0.80$ s; the visit to $-5.00$ cm at $t=0.40$ s is half a period, not a whole one
$$\omega = \sqrt{\frac{k}{m}} \;\Rightarrow\; m = \frac{k}{\omega^{2}}$$
the only unmeasured quantity, so the only one the relation can be solved for
$$m = \frac{40.0}{61.685} = 0.648\ \mathrm{kg}$$
a bit under two thirds of a kilogram
(d) The function, with the phase read off the first row
$$x(0) = +A \;\Rightarrow\; \varphi = 0$$
the table starts at the far end, which is the released from rest case
$$x = (0.0500\ \mathrm{m})\cos(7.854\,t)$$
and this reproduces every row of the table, which is the check below
Answer $$\boxed{\;A = 5.00\ \mathrm{cm},\ T = 0.800\ \mathrm{s},\ f = 1.25\ \mathrm{Hz},\ \omega = 7.85\ \mathrm{rad/s},\ m = 0.648\ \mathrm{kg}\;}$$
Check
Independent check by testing the answer against a row that was not used to build it: at $t = 0.10$ s the formula gives $5.00\cos(0.7854) = 5.00(0.7071) = 3.54$ cm, which is the recorded value; and at $t = 0.30$ s it gives $5.00\cos(2.356) = -3.54$ cm, also recorded. A wrong period would fail both.
Any table of an oscillation gives up its amplitude and its period without a formula. Everything else in the question is then a substitution.
D · interleaved 4 questions
1§14.4 — a block that arrives, then oscillates●●●●○
A 0.600 kg block slides along a horizontal floor at 2.00 m/s towards a spring. It first crosses a rough patch 0.500 m long with a coefficient of kinetic friction of 0.150, then reaches a smooth section where it meets and sticks to the free end of a spring of force constant 150 N/m.
Given
$m = 0.600\ \mathrm{kg}$, arriving at $2.00\ \mathrm{m/s}$
rough patch 0.500 m long, $\mu_{k} = 0.150$
the section with the spring is smooth
$k = 150\ \mathrm{N/m}$
$g = 9.80\ \mathrm{m/s^{2}}$
Find
(a) Find the speed of the block as it reaches the spring.
(b) Find the amplitude of the oscillation that follows.
(c) Find its period.
Hint 1/4
This is two problems joined end to end, and the join is one number. Decide what the block carries into the spring, and only then start thinking about oscillation.
Hint 2/4
On the rough patch the total falls by $f_{k}d$ with $f_{k} = \mu_{k}mg$. On the smooth part $\tfrac12 mv^{2} = \tfrac12 kA^{2}$ gives the amplitude, and $T = 2\pi\sqrt{m/k}$ gives the period.
Hint 3/4
The numbers again: $m = 0.600$ kg arriving at 2.00 m/s, a rough patch 0.500 m long with $\mu_{k} = 0.150$, then $k = 150$ N/m on a smooth floor.
Hint 4/4
The block reaches the spring at 1.59 m/s, the amplitude is 10.1 cm, and the period is 0.397 s.
Show solution
Energy for the rough patch rather than forces and kinematics, because the question links a speed to a distance and never mentions a time; and then energy again for the amplitude, for the same reason.
the friction is behind the block now, and in any case the period never depended on how much energy arrived
Answer $$\boxed{\;v = 1.59\ \mathrm{m/s},\qquad A = 0.101\ \mathrm{m},\qquad T = 0.397\ \mathrm{s}\;}$$
Check
Independent check on (b) through the speed rather than the energy: $A = v/\omega$ with $\omega = \sqrt{150/0.600} = 15.81$ rad/s, giving $A = 1.5906/15.811 = 0.1006$ m, the same to three figures and computed without ever writing down a joule.
Interleaved questions are usually built this way: an earlier section decides one number, and this section takes that number as its starting condition.
2§14.4 — something arrives and stays●●●●●
A 10.0 g pellet travelling horizontally at 300 m/s embeds itself in a 990 g block that rests at the equilibrium position on a smooth surface, attached to a spring of force constant 200 N/m. The collision is over before the block has moved appreciably.
Given
pellet $m = 0.0100\ \mathrm{kg}$ at $300\ \mathrm{m/s}$
block $M = 0.990\ \mathrm{kg}$, at rest at the equilibrium position
$k = 200\ \mathrm{N/m}$, smooth surface
the pellet stays inside the block
Find
(a) Find the speed of the block and pellet immediately after the collision.
(b) Find the amplitude and the period of the oscillation that follows.
(c) Find what fraction of the pellet's kinetic energy survives the collision.
Hint 1/4
The first thing to decide is which conservation law applies to the collision itself. Getting that wrong makes every later number wrong, and the words the pellet stays inside are the clue.
Hint 2/4
In a collision where the two stick together, momentum is conserved and kinetic energy is not: $mv = (m+M)V$. After that, the smooth spring conserves energy: $\tfrac12(m+M)V^{2} = \tfrac12 kA^{2}$, and $T = 2\pi\sqrt{(m+M)/k}$.
Hint 3/4
The numbers again: a 0.0100 kg pellet at 300 m/s into a 0.990 kg block, total mass 1.000 kg, on a spring with $k = 200$ N/m.
Hint 4/4
$V = 3.00$ m/s, $A = 0.212$ m and $T = 0.444$ s, and only 1.00% of the kinetic energy survives.
Show solution
Momentum for the collision and energy for everything after it. Mixing the two, by conserving energy through the collision, would give an amplitude nearly ten times too large.
(a) The collision: momentum, because the two stick together
$$mv = (m+M)V$$
kinetic energy is not conserved when bodies stick, so using it here would be the one fatal error in this question
ninety nine per cent of it went into deforming and heating, not into the spring
Answer $$\boxed{\;V = 3.00\ \mathrm{m/s},\qquad A = 0.212\ \mathrm{m},\qquad T = 0.444\ \mathrm{s},\qquad 1.00\%\ \text{survives}\;}$$
Check
Independent check on the surviving fraction without any numbers: for a body of mass $m$ sticking to one of mass $M$ initially at rest, the surviving fraction is $m/(m+M)$, which here is $0.0100/1.000 = 1.00\%$. The general result and the arithmetic agree, and it also explains why the answer looked suspiciously round.
Whenever something arrives and stays, the surviving fraction of kinetic energy is the mass ratio $m/(m+M)$. A light fast thing hitting a heavy slow thing wastes almost everything.
3§14.6 — two ways to use the same string●●●●○
A small bob on a light string of length 0.500 m is used in two experiments. In the first it is whirled so that it moves in a horizontal circle with the string making a constant angle of 25.0 degrees with the vertical. In the second the same bob on the same string is allowed to swing back and forth through a few degrees.
Given
$L = 0.500\ \mathrm{m}$ in both experiments
first experiment: a horizontal circle, string at $25.0^{\circ}$ to the vertical
second experiment: small back and forth swings
$\cos 25.0^{\circ} = 0.90631$
$g = 9.80\ \mathrm{m/s^{2}}$
Find
(a) Find the time for one revolution in the first experiment.
(b) Find the period in the second experiment.
(c) State which of the two is simple harmonic motion, and why the other is not.
Hint 1/4
The two experiments use the same hardware and are not the same physics. Decide for each one what is going round or going back and forth, and which law you have for it, before writing anything.
Hint 2/4
For the horizontal circle: the vertical component of the tension carries the weight and the horizontal component supplies $m\omega^{2}r$, which gives $T = 2\pi\sqrt{L\cos\theta/g}$. For the small swing: $T = 2\pi\sqrt{L/g}$.
Hint 3/4
The numbers again: $L = 0.500$ m in both, the cone angle is $25.0^{\circ}$ with $\cos 25.0^{\circ} = 0.90631$, and $g = 9.80\ \mathrm{m/s^{2}}$.
Hint 4/4
The revolution takes 1.35 s, the swing takes 1.42 s, and only the second is simple harmonic motion.
Show solution
Forces for the cone and the harmonic formula for the swing. The temptation is to use the pendulum period for both, since the apparatus is identical, and the identical apparatus is exactly the trap.
(a) The cone: forces first, because nothing oscillates
$$F_{T}\cos\theta = mg, \qquad F_{T}\sin\theta = m\omega^{2}r,\quad r = L\sin\theta$$
two components of one tension doing two different jobs, which is the circular motion argument from the earlier section
$$\omega^{2} = \frac{g}{L\cos\theta} \;\Rightarrow\; T = 2\pi\sqrt{\frac{L\cos\theta}{g}}$$
dividing the two equations removes the tension and the mass at the same time
there is no displacement from an equilibrium anywhere in the description, so the definition cannot even be applied
$$\text{cone: } T \ \text{depends on } \theta$$
and a harmonic period never depends on the size of the motion, which settles it
Answer $$\boxed{\;T_{\rm cone} = 1.35\ \mathrm{s},\qquad T_{\rm swing} = 1.42\ \mathrm{s},\ \text{only the swing is harmonic}\;}$$
Check
Independent check on (a) by a limit: as the cone angle goes to zero the formula gives $2\pi\sqrt{L/g}$, exactly the swinging period. That is no coincidence, since a very small circle is two very small swings at right angles, and it checks that the cosine is in the right place.
Identical hardware does not mean identical physics. What decides which formulas apply is the shape of the net force, never the equipment list.
4§14.6 — two pivots that keep the same time●●●●●
A uniform rod of length 1.20 m can be hung from a pivot and swung through a small angle in a vertical plane. Two pivots are considered: one at the very end of the rod, and one at a point a sixth of the length from the centre, that is 0.200 m from the centre.
Given
uniform rod, $L = 1.20\ \mathrm{m}$, mass $M$
$I_{\rm cm} = \tfrac{1}{12}ML^{2}$
pivot 1: at the end, so $d_{1} = L/2 = 0.600\ \mathrm{m}$
pivot 2: at $d_{2} = L/6 = 0.200\ \mathrm{m}$ from the centre
$g = 9.80\ \mathrm{m/s^{2}}$, small swings
Find
(a) Find the period for the pivot at the end.
(b) Find the period for the pivot a sixth of the length from the centre.
(c) Comment on the comparison and give the physical reason for it.
Hint 1/4
Both parts are the same formula used twice, and the interesting work is in getting the moment of inertia about each pivot before substituting anything. Do not expect the two answers to be different just because the pivots are.
Hint 2/4
$T = 2\pi\sqrt{I/(Mgd)}$ with $I = I_{\rm cm} + Md^{2}$ from the parallel axis theorem, and $d$ the distance from the pivot to the centre of mass.
Hint 3/4
The numbers again: $L = 1.20$ m, $I_{\rm cm} = \tfrac{1}{12}ML^{2}$, and the two distances are $d_{1} = 0.600$ m and $d_{2} = 0.200$ m.
Hint 4/4
Both periods come out at 1.80 s: the equivalent length $I/Md$ is $2L/3 = 0.800$ m for both pivots.
Show solution
Computing the equivalent length rather than the period twice, because the mass and the $2\pi$ cancel out of it and the comparison becomes a single number.
Independent check that this is a real pattern rather than a coincidence of these numbers: the expression $L_{\rm eq} = I_{\rm cm}/(Md) + d$ takes the same value at $d$ and at $I_{\rm cm}/(Md)$, because swapping those two swaps the two terms. For a rod, $I_{\rm cm}/M = L^{2}/12$, so the partner of $d = L/2$ is $(L^{2}/12)/(L/2) = L/6$, exactly the second pivot. The pairing is general and works for any shape.
Every physical pendulum has a partner pivot on the other side of the centre of mass that keeps the same time, which is the basis of a classical way of measuring $g$ with a rod and a stopwatch.
Mistake ledger (22 entries)
⚠ Measuring the displacement of a hanging mass from the unstretched length
The natural length is the one the spring has when it is lying on the bench, so it feels like the honest zero, and the resting position looks like an accident of the mass that happens to be hanging there.
The sign carries no number, so it is the first thing to be lost when the equation is copied, and every subsequent line still looks tidy.
wrong$$a = +\frac{k}{m}x$$
right$$a = -\frac{k}{m}x$$
⚠ Putting the frequency in hertz where the angular frequency belongs
Both are called frequency in ordinary speech and both come out of the same calculation, so the factor of $2\pi$ between them has nowhere obvious to live.
They differ by a single factor of $\omega$ and the two formulas sit next to each other on every formula sheet, so the eye picks the wrong one under pressure.
⚠ Taking the magnitude and losing the direction of the acceleration
The formula $a=-\omega^{2}x$ looks like a size, and the minus sign is easy to read as decoration rather than as the statement that the acceleration always points at the middle.
wrong$$a = +\omega^{2}x \Rightarrow \text{acceleration points away from the middle}$$
right$$a = -\omega^{2}x \Rightarrow \text{acceleration always points at the middle}$$
⚠ Using the position form of the speed and then asking which way the block is going
The expression $v = \pm\omega\sqrt{A^{2}-x^{2}}$ answers so much with so little that it feels like it must contain everything.
wrong$$v = +\omega\sqrt{A^{2}-x^{2}}\ \text{, so the block moves in the } +x \text{ direction}$$
right$$|v| = \omega\sqrt{A^{2}-x^{2}},\ \text{direction from } v=-A\omega\sin(\omega t+\varphi)$$
⚠ Putting centimetres into the energy formula
Amplitudes are quoted in centimetres because that is what they look like in a laboratory, and the number goes straight into the formula without a stop.
⚠ Assuming the energy is shared equally at half the amplitude
Half the distance sounds like half the energy, and the square in the store is invisible until you write it down.
wrong$$x = \tfrac12 A \;\Rightarrow\; U = \tfrac12 E$$
right$$x = \tfrac12 A \;\Rightarrow\; U = \tfrac14 E, \qquad U = \tfrac12 E \text{ at } x = A/\sqrt2$$
⚠ Believing the block itself is going round something
The formula is full of angles and the symbol is called angular frequency, so the mind supplies a rotation that is not there.
wrong$$\text{block on a spring: } v = \omega r \text{ with } r \text{ the amplitude, at all times}$$
right$$\text{block on a spring: } |v| = A\omega \text{ only at } x=0,\ \text{and } v=0 \text{ at } x=\pm A$$
⚠ Taking the phase constant to be in degrees because it was found with an inverse tangent
Calculators hand back inverse trigonometric results in whatever mode they are in, and a phase of 39.8 looks as reasonable as a phase of 0.695.
wrong$$x = A\cos(20.0\,t + 39.8)$$
right$$x = A\cos(20.0\,t + 0.695)$$
⚠ Feeding degrees into the small angle step
The amplitude of a swing is naturally described in degrees, and the approximation $\sin\theta\approx\theta$ is then applied to the number that is written down.
⚠ Expecting the amplitude at resonance to be infinite
The undamped expression has a zero in the denominator at the natural frequency, and it is tempting to stop reading there.
wrong$$\omega_{\rm drive} \to \omega_{0} \;\Rightarrow\; A \to \infty$$
right$$\omega_{\rm drive} \to \omega_{0} \;\Rightarrow\; A \to \frac{F_{0}}{b\,\omega_{0}}$$
⚠ Using a constant acceleration formula on an oscillation
The formulas of the kinematics sections are the ones that come to hand first, and nothing in the problem shouts that the acceleration is changing. The error is always 10.0% low, for every spring and every pull.
⚠ Leaving the calculator in degree mode inside an inverse cosine
The inverse cosine returns whatever the calculator is set to, and 120 is such a familiar angle that it does not look like a unit at all. Comparing the answer with the period catches it instantly: 6.33 s is nineteen periods.
wrong$$\omega t = \cos^{-1}(-0.500) = 120 \;\Rightarrow\; t = \frac{120}{18.97} = 6.33\ \mathrm{s}$$
right$$\omega t = 2.094\ \mathrm{rad} \;\Rightarrow\; t = \frac{2.094}{18.97} = 0.110\ \mathrm{s}$$
⚠ Reading a period off a table as the gap between the two extremes
The two extremes are the most visible features of the data, and the trip between them is only half a cycle. A period is a return to the same state, which includes the direction of travel.
right$$T = 0.800\ \mathrm{s}\ \text{(back to } +5.00\ \mathrm{cm}\ \text{moving the same way)}$$
⚠ Conserving kinetic energy through a collision in which the bodies stick
Energy is the tool that has just been used for the oscillation, so the hand reaches for it again at the collision, where it does not hold. The surviving fraction here is only 1.00%.
wrong$$\tfrac12 m v^{2} = \tfrac12 (m+M)V^{2} \;\Rightarrow\; V = 30.0\ \mathrm{m/s}$$
right$$m v = (m+M)V \;\Rightarrow\; V = 3.00\ \mathrm{m/s}$$
⚠ Using the friction force with the wrong distance
Every other term in an energy equation depends only on where the body started and finished, so friction gets treated the same way. It is the one term that counts the whole journey.
wrong$$W_{f} = f_{k}\times(\text{straight line between the ends})$$
right$$W_{f} = f_{k}\times(\text{distance actually travelled along the surface})$$
⚠ Applying the simple pendulum formula to a whirled bob
The apparatus is identical, a bob on a string, and the apparatus is what the eye recognises. What decides the formula is the shape of the net force, and a conical pendulum has no restoring force at all.
steady driving, after the starting transient; the peak height is set by the damping
Check yourself
Close the page and write, from memory: the condition a force must satisfy for simple harmonic motion; the position, velocity and acceleration in time with the place where each maximum occurs; the angular frequency of a mass on a spring and of a simple pendulum; the total energy and the speed at a given position. Then open the formula card and mark what you missed rather than what you got.
Find a system's equilibrium position, show the net force is $-k_{\rm eff}x$ and read off the effective stiffness, including for a hanging mass?
c-restoring-force
Write $x(t)$ for an oscillator, get its period from the mass and the stiffness, and fix the amplitude and the phase constant from a position and a velocity at one instant?
c-shm-solution
Produce the velocity and the acceleration at a named instant or a named position, and say without thinking where each of them is largest and where each is zero?
c-velocity-acceleration
Get a speed from a position, or an amplitude from an arriving speed, using $\tfrac12 kA^{2}$ and without ever writing down a phase constant?
c-energy-shm
Draw the reference circle for a given motion and point at the amplitude, the angular frequency and the phase constant on it?
c-circular-shadow
Derive the pendulum period from the tangential weight, say what the costs, and handle a rigid body with the parallel axis theorem?
c-pendulum
Say how many cycles a lightly damped oscillator survives, explain why its frequency barely changes, and locate the driving rate at which a system responds most?
c-damping-resonance
Glossary (22 terms)
simple harmonic motionbasit harmonik hareket
The motion of a body whose net force is proportional to its displacement from one position and directed back towards it. Its position in time is a cosine and its period does not depend on the size of the motion.
restoring forcegeri çağırıcı kuvvet
A force that always points back towards the equilibrium position, from either side of it. Without one there is no oscillation, and if it is not proportional to the displacement the oscillation is not the simple kind.
equilibrium positiondenge konumu
The one place where the net force on the body is zero, and the place from which every displacement in this section is measured. For a hanging mass it is where the mass rests, not the free end of the unstretched spring.
amplitudegenlik
The largest displacement from the equilibrium position that the motion reaches. It is fixed by how the motion was started, never by the spring, and it never appears in the period.
periodperiyot
The time for one complete there and back trip, in seconds. For a mass on a spring it is set by the mass and the stiffness alone.
frequencyfrekans
The number of complete cycles per second, in hertz, equal to one over the period. It is not the quantity that goes inside the cosine.
hertzhertz
The unit of frequency, one cycle per second. A block at 3 Hz makes three complete there and back trips every second.
angular frequencyaçısal frekans
The rate at which the phase advances, in radians per second, equal to $2\pi$ times the frequency. It is the quantity that appears inside the cosine and it is what $\sqrt{k/m}$ produces.
phasefaz
The whole angle inside the cosine at a given instant, $\omega t + \varphi$, in radians. It advances by $2\pi$ every period, which is what makes the motion repeat.
phase constantfaz sabiti
The value of the phase at the instant the clock is started. On the reference circle it is the angle the point was sitting at, which is why it is in radians and why adding a whole turn changes nothing.
force constantyay sabiti
The constant $k$ in $F=-kx$, in newtons per metre, measuring how hard a spring pulls back per metre of stretch. A stiff spring has a large one.
Hooke's lawHooke yasası
The statement that an ideal spring pulls back in proportion to its stretch, $F = -kx$. It is an approximation that holds while the spring is not stretched too far, and everything in this section rests on it.
effective force constant
The constant read off the net force of a system that is not a plain spring, once that force has been written as minus a constant times the displacement. For a pendulum it is $mg/L$, which is why the mass cancels.
reference circle
The circle of radius equal to the amplitude on which an imaginary point travels at a steady rate, so that its shadow on a diameter performs the oscillation. It gives all three constants their meaning in one picture.
turning pointdönüm noktası
A position at which the body is momentarily at rest and reverses. In an oscillation these are the two ends of the swing, where the whole of the energy is stored and the acceleration is largest.
simple pendulumbasit sarkaç
A small bob on a light inextensible string. For small swings its period is $2\pi\sqrt{L/g}$, independent of the mass of the bob and of the size of the swing.
physical pendulumfiziksel sarkaç
Any rigid body swinging about a pivot that is not its centre of mass, with period $2\pi\sqrt{I/mgd}$ and the moment of inertia taken about the pivot.
small angle approximationküçük açı yaklaşımı
Replacing $\sin\theta$ by $\theta$ for an angle measured in radians. It is what turns a pendulum into a harmonic oscillator, and it costs about 0.4% in the period at a swing of fifteen degrees.
damped harmonic motionsönümlü harmonik hareket
Oscillation in the presence of a drag force proportional to the velocity. The amplitude falls exponentially while the frequency is almost untouched, provided the damping is light.
kritik sönüm
The amount of damping, $b = 2\sqrt{mk}$, that returns a displaced system to equilibrium in the shortest time without overshooting. Beyond it the system creeps back and never oscillates.
natural frequencydoğal frekans
The frequency at which a system oscillates when it is displaced and left alone. For a mass on a spring it is $\sqrt{k/m}$ divided by $2\pi$.
resonancerezonans
The large response of a system that is driven at or very near its natural frequency. The height of the peak is limited only by the damping, which is why lightly damped structures are the dangerous ones.
What comes next
That is the end of the course. The natural next step is to put many oscillators side by side and couple each one to its neighbours: each still obeys the law of this section, but the motion no longer stays in one place, it travels along the line of them. That travelling oscillation is a wave, and it is what the next physics course opens with.
Sources
D. Giancoli, Physics for Scientists and Engineers with Modern Physics, 5th edition The set textbook. Its chapter on oscillations covers the same ground as this section and in a similar order, and its end of chapter problems are the right next step once this practice set feels comfortable.
Course syllabus, week 14 line and assessment table The scope comes from the week line, which reads Oscillations and quotes no chapter numbers, so no chapter number is quoted here either. The weightings on the summary card come from the assessment table and nothing beyond them is claimed.
SI units and the constants used here Displacements in metres, periods in seconds, frequency in hertz, angular frequency in radians per second, force constants in newtons per metre and damping constants in newton seconds per metre. Throughout, $g = 9.80\ \mathrm{m/s^{2}}$, the same value as in every earlier section.