What is the difference between and f?

f counts cycles per second, in hertz, and = 2 f counts radians per second. The relations v = A, a = A2 and a = -2 x all use .

Unit 7 · Topic 7.3 · about 30 minutes · 7 parts · 12 scored checks, 2 of them multiple choice, plus 2 from earlier lessons

AP Physics C: Mechanics 7.3: Representing and Analyzing SHM

Q: Why must the angle be in radians?

Because is in radians per second, t is in radians, and the derivatives vx = -A( t + ) and ax = -A2( t + ) are only right in radians. Set your calculator to radian mode.

Built by Tanner Crow from College Board's course framework · Every answer key independently checked

Unit 7: 10 to 15% of the multiple-choice section

What you'll be able to do

Describe the displacement, velocity and acceleration of an object in simple harmonic motion, from its position function, from its graphs or from measured data.

Key vocabulary

Term	What it means	The exam trap
Simple harmonic motion (SHM)	Periodic motion in which the restoring force, and so the acceleration, is proportional to the displacement from equilibrium and opposite to it.	Not every repeating motion is SHM: a ball bouncing on the floor repeats, but the force on it is not proportional to its displacement.
Equilibrium position	The position at which the net force on the oscillating object is zero; SHM displacements are measured from it.	A sensor or a ruler may put its zero somewhere else, so subtract the equilibrium reading before you treat a reading as a displacement.
Amplitude	The largest displacement from the equilibrium position, $A$ in $x = A \cos (ω t + ϕ)$ .	On a graph, read it as a peak's height above the equilibrium line, which is not the line $x = 0$ when positions are measured from somewhere other than equilibrium.
Frequency	The number of complete cycles per second, $f = \frac{1}{T}$ , in hertz.	A frequency in hertz and an angular frequency in radians per second are different numbers for the same motion, so check which one an equation calls for.
Angular frequency	The rate at which the argument of the cosine grows, $ω = 2 π f = \frac{2 π}{T}$ , in radians per second.	$ω$ is not a speed of the object; the object's speed changes all through the cycle while $ω$ stays fixed.
Phase constant	The angle $ϕ$ in $x = A \cos (ω t + ϕ)$ , in radians, that sets where in its cycle the object is at $t = 0$ .	With $ω t$ in radians, $ϕ$ must be in radians too, and your calculator must be in radian mode when you evaluate the function.
Second-order differential equation	An equation that relates a function to its second derivative, here $\frac{d^{2} x}{d t^{2}} = - ω^{2} x$ , from Newton's second law with a restoring force.	Its solution is a whole function, $x (t)$ , not one number; you are expected to know that solution, not to prove it.
Natural frequency	The frequency at which a system oscillates when it is displaced from its equilibrium position and let go.	It belongs to the system, its masses and springs, not to whatever is pushing on it.
Resonance	The growth in amplitude that occurs when a sinusoidal external force is exerted at the natural frequency of an oscillating system.	Resonance is about timing, so name both frequencies, the driving one and the natural one, before deciding whether it happens.

A graph of position x in meters, measured from equilibrium with right positive, against time t in seconds from 0 to 6 s, for x(t) = 0.15 sin(pi t / 2). The curve starts at 0, peaks at 0.15 m at 1 s, crosses 0 at 2 s, reaches negative 0.15 m at 3 s, crosses 0 at 4 s and peaks again at 5 s. Dashed lines at 0.15 m and negative 0.15 m are labeled A = 0.15 m and negative A, and a bracket from 1 s to 5 s is labeled T = 4.0 s. A flat line touching the first peak is labeled slope 0, and a tangent line at 2 s is labeled slope negative 0.24 m/s.

Part 1 Think first Pull it farther

Commit to a guess first. Being wrong now makes the right idea stick better.

Predict first

A teacher attaches a block to a spring on a level air track, where friction is negligible. She pulls the block 2 cm from where it rests, lets go, and times 10 full oscillations. Then she pulls it 4 cm, lets go, and times 10 more. How do the two times compare?

In short. In simple harmonic motion the displacement from equilibrium is $x = A \cos (ω t + ϕ)$ , the solution of $\frac{d^{2} x}{d t^{2}} = - ω^{2} x$ ; its derivatives give the velocity and an acceleration $a = - ω^{2} x$ , with largest values $v_{\max} = A ω$ and $a_{\max} = A ω^{2}$ .

At equilibrium the speed is greatest and the acceleration is zero; at a turning point the velocity is zero and the acceleration is largest, pointing back toward equilibrium.
The period, $T = \frac{2 π}{ω}$ , is set by the system, so changing the amplitude does not change it.
A sinusoidal external force at the system's natural frequency produces resonance, which increases the amplitude.

Part 2 Learn it Four ideas, with a check after each

3a. The equation, and the function that solves it

In Topic 7.1 an ideal spring exerted a restoring force $F_{x} = - k x$ , with $x$ measured from the equilibrium position. Newton's second law, with $a_{x} = \frac{d^{2} x}{d t^{2}}$ , turns that force into an equation about the motion:

m \frac{d^{2} x}{d t^{2}} = - k x ⟹ \frac{d^{2} x}{d t^{2}} = - ω^{2} x

with $ω^{2} = \frac{k}{m}$ for a block on a spring. This is a second-order differential equation: it ties the position to its own second derivative. Any system whose acceleration is proportional to its displacement and opposite to it obeys it, and that is how you identify SHM.

The position as a function of time is its solution, which you are expected to know but not to prove:

x = A \cos (ω t + ϕ)

Here $A$ is the amplitude, the largest displacement from equilibrium, and $ω = 2 π f = \frac{2 π}{T}$ is the angular frequency in rad/s. The phase constant $ϕ$ , in radians, says where in the cycle the object is at $t = 0$ .

Two starts are common. Released from rest at $x = + A$ , an object has $ϕ = 0$ , so $x = A \cos (2 π f t)$ . Passing through equilibrium at $t = 0$ moving in the $+ x$ direction, it has $x = A \sin (2 π f t)$ , which is the cosine with $ϕ = - \frac{π}{2}$ .

Quick check · Fill the statement

A glider on a spring rests at its equilibrium position, $x = 0$ , with right positive. A student pulls it 0.05 m to the right and releases it from rest at $t = 0$ . It then completes 4 full cycles in 2.0 s.

The glider's amplitude is , its frequency is , and its position as a function of time is .

3b. Velocity and acceleration from the position

Velocity and acceleration come from the position function by differentiating, with the chain rule on the cosine:

v_{x} = \frac{d x}{d t} = - A ω \sin (ω t + ϕ) a_{x} = \frac{d v_{x}}{d t} = - A ω^{2} \cos (ω t + ϕ)

Comparing with $x$ , the acceleration is $- ω^{2}$ times the position at the same instant:

a = - ω^{2} x

So the acceleration always points toward equilibrium, and its size grows with the distance from equilibrium.

A sine or cosine is never larger than 1 in size, so the largest speed and the largest size of the acceleration are

v_{\max} = A ω a_{\max} = A ω^{2}

For a block with $A = 0.050$ m and $ω = 8.0$ rad/s, $v_{\max} = 0.40$ m/s and $a_{\max} = 3.2$ m/s². When that block is at $x = + 0.020$ m, its acceleration is $a = - (8.0)^{2} (0.020) = - 1.28$ m/s², pointing back toward equilibrium.

Quick check · Calculate

A block oscillates on a horizontal spring with angular frequency 6.0 rad/s and amplitude 0.050 m about its equilibrium position at $x = 0$ , with right positive. At one instant the block is 0.030 m to the left of equilibrium. What is its acceleration at that instant?

Your answer m/s²

3c. Zeros, peaks and the period

The position, velocity and acceleration graphs are sinusoids with the same period, each a quarter cycle from the next. Their zeros and extremes line up like this:

Where each quantity is zero or largest, for $x = A \cos (ω t + ϕ)$ with right positive
Place in the cycle	$x$	$v_{x}$	$a_{x}$
Turning point on the right	$+ A$	0	$- A ω^{2}$
Passing equilibrium, moving left	0	$- A ω$	0
Turning point on the left	$- A$	0	$+ A ω^{2}$
Passing equilibrium, moving right	0	$+ A ω$	0

On a graph of $x$ against $t$ the velocity is the slope: zero at every peak and trough, steepest where the curve crosses equilibrium, as the graph at the top of the page shows. Read $A$ as a peak's height above the equilibrium line, and $T$ from one peak to the next.

The period is $T = \frac{2 π}{ω}$ , and $ω$ comes from the system itself, $\sqrt{k / m}$ for a block on a spring. The amplitude appears nowhere in it. Pull the block twice as far and it has twice as far to go, but $v_{\max} = A ω$ doubles too, so each cycle takes the same time.

Quick check · Decide and justify

A glider on a horizontal spring has position $x (t) = 0.12 \sin (π t)$ , with $x$ in meters, $t$ in seconds and the angle in radians. Its equilibrium position is $x = 0$ , and right is positive. At $t = 0.50$ s, what is the size of the glider's acceleration?

Which reason supports your answer?

3d. Natural frequency and resonance

Every oscillating system has a natural frequency: the frequency at which it oscillates when it is displaced from its equilibrium position and let go. For a block on a spring that is $f = \frac{1}{2 π} \sqrt{\frac{k}{m}}$ .

Now let a sinusoidal external force push on the system. When the force is exerted at the natural frequency, each push arrives at the same point of every cycle and adds to the motion, and the amplitude builds up. That is resonance: it occurs when an external force is exerted at the natural frequency of an oscillating system, and it increases the amplitude of the motion.

Pushing a child on a swing works this way. You push once per swing, at the swing's own frequency, and each push adds a little. Push at some other frequency and your pushes drift out of step with the swing: some help and some oppose, and the swing never gets high.

Engineers design around it. A footbridge whose natural frequency is close to the rate at which people step can sway badly under a crowd, and a glass driven by a loud sound at its natural frequency can crack.

Quick check · Multiple choice

A 0.50 kg cart on a level track is attached to one end of a spring with spring constant 200 N/m. A motor moves the spring's other end back and forth sinusoidally, at a frequency a student can set. At which driving frequency does the cart's amplitude grow largest?

Common mistakes

The mistake	Why it is wrong	Do this instead
Taking the distance between the two turning points as the amplitude.	The amplitude is measured from the equilibrium position. The turning points are $+ A$ and $- A$ , so they are $2 A$ apart.	Read the amplitude as the distance from equilibrium to one turning point, or half the distance between the two.
Expecting the acceleration to be zero wherever the velocity is zero.	In SHM $a = - ω^{2} x$ . At a turning point the velocity is zero but $\| x \| = A$ , so the acceleration there is as large as it ever gets.	Find the acceleration from the position, $a = - ω^{2} x$ , never from the size of the velocity.
Expecting a larger amplitude to give a longer period.	$T = \frac{2 π}{ω}$ , and $ω$ depends only on the system, such as $k$ and $m$ for a spring. A larger amplitude means a longer trip at a proportionally higher speed.	Change the period by changing the system, not by pulling it farther; use the amplitude for $v_{\max}$ and $a_{\max}$ , which do scale with it.

Part 3 Watch it done A block passing through equilibrium

Watch each step, then do the last one yourself.

A 0.40 kg block on a level, frictionless surface is attached to a spring with spring constant 160 N/m. Its equilibrium position is $x = 0$ , and right is positive. At $t = 0$ the block passes through equilibrium moving right at 1.2 m/s. Write the block's position as a function of time, and find its largest acceleration.

Step 1 · Find the angular frequency
For a block on a spring, $ω = \sqrt{\frac{k}{m}} = \sqrt{\frac{160}{0.40}} = \sqrt{400} = 20$ rad/s.
Step 2 · Find the amplitude
At equilibrium the block moves at its largest speed, so $v_{\max} = A ω = 1.2$ m/s and $A = \frac{1.2}{20} = 0.060$ m.
Why this step?
Why is 1.2 m/s the largest speed, and not just the speed at $t = 0$ ?
At equilibrium $x = 0$ , so $a = - ω^{2} x = 0$ : the velocity has stopped growing and has not yet begun to shrink. On the way in toward equilibrium the spring speeds the block up, and on the way out it slows the block down, so the speed peaks at equilibrium.
Step 3 · Choose the phase
At $t = 0$ , $x = 0$ and the block moves in the $+ x$ direction, which the sine gives: $x (t) = 0.060 \sin (20 t)$ m, the same as $0.060 \cos (20 t - \frac{π}{2})$ m.
Your step · Calculate

Finish it. What is the largest size of the block's acceleration? Right is positive.

Your answer m/s²

Part 4 Try it New situations

Try it · Sketch the graph

A block on a horizontal spring oscillates with a period of 2.0 s about its equilibrium position, $x = 0$ , with right positive. It is released from rest at $x = + 10$ cm at $t = 0$ . Sketch the shape of its velocity $v_{x}$ against $t$ from 0 to 0.50 s: drag the points so the curve rises or falls, and bends or runs straight, the way $v_{x}$ does.

Drag a point up or down, or tap it and use the arrow keys. Only the shape is checked.

Try it · Calculate

A glider on a level air track oscillates about its equilibrium position, $x = 0$ , with right positive. Its position is $x (t) = 0.040 \cos (5.0 t + \frac{π}{3})$ , with $x$ in meters, $t$ in seconds and the angle in radians. What is the glider's velocity at $t = 0$ ?

Your answer m/s

On paper. This time the glider starts on the left, at $x = - A$ , not at $+ A$ as in the items above. It is released from rest there at $t = 0$ , with right positive. Sketch its position, velocity and acceleration against time for one full period on three sets of axes stacked one above the other, so the same instant lines up on all three. Mark $\frac{T}{4}$ , $\frac{T}{2}$ , $\frac{3 T}{4}$ and $T$ on each time axis, and label each graph's largest value in terms of $A$ and $ω$ . There is nothing to submit here; once you open the Explain it model, check that each of your three graphs is that model's graph turned upside down.

Part 5 Explain it Draw it, derive it, graph it

On the exam this is handwritten, so answer on paper first. Nothing you type here leaves this page. Then check your answer against the model and each scoring point.

Free response is 50% of the exam score. This one is shaped like Question 2, Translation Between Representations.

Free-response practice · Question 2 style

A glider of mass $m$ on a level, frictionless air track is attached to a spring with spring constant $k$ , whose other end is fixed. The glider's equilibrium position is $x = 0$ , and right is positive. The glider is pulled to $x = + A$ and released from rest at $t = 0$ .

(a) On paper, draw a free-body diagram of the glider at the instant it is released. Draw each force as an arrow starting on a dot, and label it.

(b) Starting from Newton's second law, derive the differential equation for the glider's position $x$ . Then write $x (t)$ in terms of $A$ , $k$ , $m$ and $t$ .

(c) Derive expressions for the glider's largest speed and the largest size of its acceleration, in terms of $A$ , $k$ and $m$ .

(d) On paper, sketch graphs of $x$ , $v_{x}$ and $a_{x}$ against $t$ from $t = 0$ to $t = T$ , one above the other with their time axes lined up. Label the time axis at $\frac{T}{4}$ , $\frac{T}{2}$ , $\frac{3 T}{4}$ and $T$ , and label each graph's largest value.

(e) The glider is now pulled to $x = + 2 A$ and released from rest. Predict whether the period, the largest speed and the largest acceleration each increase, decrease or stay the same, and justify each prediction using your answers to (b) and (c).

Your answer stays on this page. It is never sent or saved.

Model answer

(a) A dot with three arrows: a downward gravitational force exerted by Earth, an upward normal force exerted by the track, the same length, and a spring force of size $k A$ pointing left, toward equilibrium.

(b) The spring exerts $F_{x} = - k x$ , so Newton's second law gives $m \frac{d^{2} x}{d t^{2}} = - k x$ , or $\frac{d^{2} x}{d t^{2}} = - \frac{k}{m} x$ . That is $\frac{d^{2} x}{d t^{2}} = - ω^{2} x$ with $ω = \sqrt{\frac{k}{m}}$ . Released from rest at $+ A$ , the glider has $ϕ = 0$ , so $x (t) = A \cos (\sqrt{\frac{k}{m}} t)$ .

(c) $v_{x} = \frac{d x}{d t} = - A \sqrt{\frac{k}{m}} \sin (\sqrt{\frac{k}{m}} t)$ , whose largest size is $v_{\max} = A \sqrt{\frac{k}{m}}$ . Then $a_{x} = \frac{d v_{x}}{d t} = - \frac{k A}{m} \cos (\sqrt{\frac{k}{m}} t)$ , so $a_{\max} = \frac{k A}{m}$ .

(d) Position: a cosine from $+ A$ , zero at $\frac{T}{4}$ , $- A$ at $\frac{T}{2}$ , zero at $\frac{3 T}{4}$ and $+ A$ at $T$ . Velocity: zero at the start, $- A ω$ at $\frac{T}{4}$ , zero at $\frac{T}{2}$ , $+ A ω$ at $\frac{3 T}{4}$ and zero at $T$ . Acceleration: $- A ω^{2}$ at the start, zero at $\frac{T}{4}$ , $+ A ω^{2}$ at $\frac{T}{2}$ , zero at $\frac{3 T}{4}$ and $- A ω^{2}$ at $T$ . Compare your sketches with those values: the velocity is zero wherever the position has a peak or a trough, and the acceleration has the opposite sign to the position at every instant.

(e) The period stays the same: $T = \frac{2 π}{ω}$ , and $ω = \sqrt{\frac{k}{m}}$ from (b) has no amplitude in it. The largest speed and the largest acceleration both double, because both expressions in (c) are proportional to the amplitude. In (d) the zeros and peaks stay at the same times; only the heights double.

Score your own answer. This is a self-check, not a grade. Tick each point your answer makes.

Draws a free-body diagram with the gravitational force down, the normal force up and the spring force to the left, each labeled, and no other forces.
Applies Newton's second law with $F_{x} = - k x$ to get $\frac{d^{2} x}{d t^{2}} = - \frac{k}{m} x$ , identifies $ω = \sqrt{k / m}$ , and writes $x (t) = A \cos (\sqrt{k / m} t)$ for a release from rest at $+ A$ .
Differentiates $x (t)$ to get $v_{\max} = A \sqrt{k / m}$ and again to get $a_{\max} = \frac{k A}{m}$ .
Sketches lined-up graphs: position a cosine from $+ A$ ; velocity zero at 0, $\frac{T}{2}$ and $T$ with $- A ω$ at $\frac{T}{4}$ ; acceleration opposite in sign to position at every instant.
Predicts the period stays the same because $ω = \sqrt{k / m}$ has no amplitude in it, and the largest speed and acceleration double because both are proportional to the amplitude.

Part 6 Check yourself A glider and a motion sensor

The case questions use the sensor readings. Answer them from those.

A glider on a level air track is attached to a spring. Its equilibrium position is $x = 0$ , and the direction away from the motion sensor is positive. The sensor's software is set to read zero at the equilibrium position. A student pulls the glider to $x = + 6.00$ cm and releases it from rest at $t = 0$ . The sensor's readings over the next 0.80 s:

Motion sensor readings
$t$ (s)	$x$ (cm)
0	6.00
0.10	4.24
0.20	0.00
0.30	−4.24
0.40	−6.00
0.50	−4.24
0.60	0.00
0.70	4.24
0.80	6.00

Case question 1 · Calculate

Read the amplitude and the period from the readings, and use them with the SHM model to calculate the glider's largest speed.

Your answer m/s

Case question 2 · Multiple choice

Using the readings, which describes the glider's motion at $t = 0.30$ s? The direction away from the sensor is positive.

Case question 3 · Rank it

Using the readings, rank the instants $t = 0.10$ s, $0.20$ s, $0.40$ s and $0.70$ s by the size of the glider's acceleration.

Greatest first

Give each one a place, 1 for first. Give two the same place if they are equal.

0.10 s
0.20 s
0.40 s
0.70 s

Your ranking: none yet

The rest stand on their own and do not use the case.

Check yourself · Put it in order

A block of mass $m$ on a level, frictionless surface is attached to a spring with spring constant $k$ , and $x$ is measured from the equilibrium position. Put the steps in order to get the differential equation for $x$ and write its solution.

Check yourself · Spot the error

A 0.20 kg glider on a spring with spring constant 20 N/m oscillates with an amplitude of 0.05 m. A student predicts what happens when the amplitude is doubled to 0.10 m. Tap the first line with a mistake.

Part 7 Keep it Bring back earlier ideas

This lesson stands on Topic 7.2, where period, frequency and angular frequency are tied together by $ω = 2 π f = \frac{2 π}{T}$ , and on Topic 1.2: velocity and acceleration are derivatives of position, and an object speeds up only when its velocity and acceleration point the same way. Pulling them back up is what makes them last.

From 1.2 Displacement, Velocity, and Acceleration · Decide and justify

A cart moves along a track with its position given by $x (t) = 5 t^{2} - t^{3}$ , with $x$ in meters and $t$ in seconds. Take the direction of increasing $x$ as positive. At $t = 2$ s, is the cart speeding up or slowing down?

Which reason supports your answer?

From 7.2 Frequency and Period of SHM · Fill the statement

A cart attached to a spring rolls back and forth about its equilibrium position on a level track. It completes 20 full cycles in 8.0 s.

The cart's period is , its frequency is , and its angular frequency is .

In three lines

In SHM, $x = A \cos (ω t + ϕ)$ is the solution of $\frac{d^{2} x}{d t^{2}} = - ω^{2} x$ , which Newton's second law gives for a restoring force proportional to displacement.
Differentiating gives $v_{x} = - A ω \sin (ω t + ϕ)$ and $a = - ω^{2} x$ , so $v_{\max} = A ω$ at equilibrium and $a_{\max} = A ω^{2}$ at the turning points.
The period does not depend on the amplitude, and a sinusoidal force at the natural frequency produces resonance, which increases the amplitude.

Exam strategy for this topic

Section I, Multiple Choice: 42 questions in 85 minutes.
Section II, Free Response: 4 questions in 95 minutes.
This lesson's Explain it is shaped like free-response question 2, Translation Between Representations.
A four-function, scientific, or graphing calculator is allowed on both sections of the exam.

When a free-response part asks for a velocity or an acceleration in SHM, write the derivative of $x = A \cos (ω t + ϕ)$ before any numbers, and say which way is positive. When a part asks you to sketch or compare graphs, line up the instants: the velocity is zero where the position has a peak or a trough, and the acceleration has the opposite sign to the position. State the solution of the differential equation when you need it; you are not asked to prove it.

Questions students ask

Do I have to prove that

x = A \cos (ω t + ϕ)

solves the differential equation?

No. The course expects you to know that solution and to recognize SHM from its equation, not to prove the solution is correct. You do need to differentiate the solution to find the velocity and the acceleration.

Should I use sine or cosine?

Either, since they differ only by a phase constant. A cosine with $ϕ = 0$ fits an object released from rest at $+ A$ ; a sine fits one passing through equilibrium in the positive direction at $t = 0$ . Pick the one whose value and direction at $t = 0$ match the start.

Why must the angle be in radians?

Because $ω$ is in radians per second, $ω t$ is in radians, and the derivatives $v_{x} = - A ω \sin (ω t + ϕ)$ and $a_{x} = - A ω^{2} \cos (ω t + ϕ)$ are only right in radians. Set your calculator to radian mode.

What is the difference between

ω

and

f

$f$ counts cycles per second, in hertz, and $ω = 2 π f$ counts radians per second. The relations $v_{\max} = A ω$ , $a_{\max} = A ω^{2}$ and $a = - ω^{2} x$ all use $ω$ .

Is resonance always a problem?

No. A swing, a musical instrument and a radio tuner use it on purpose. It is a problem when something that should stay still, such as a bridge or a machine mount, is driven at its natural frequency.

Beyond the AP Exam

College Board sets this limit on what the AP Exam asks about this topic. Anything past it is not on the exam.

AP Physics C: Mechanics only expects students to know the solution to the second-order differential equation that describes SHM, as well as be able to identify SHM. AP Physics C: Mechanics does not expect students to mathematically prove that the solution is correct.

Course alignment, for teachers

AP Physics C: Mechanics topic 7.3, Unit 7: Oscillations.

Skill 1.C: Create qualitative sketches of graphs that represent features of a model or the behavior of the physical system.
Skill 2.A: Derive a symbolic expression from known quantities by selecting and following a logical mathematical pathway.
Skill 2.C: Compare physical quantities between two or more scenarios or at different times and/or locations within a single scenario.
Skill 3.B: Apply an appropriate law, definition, theoretical relationship, or model to make a claim.

This lesson maps all 13 essential knowledge statements of the topic to the part that teaches each one: 13 are checked by a scored question. The map is in the lesson file.