AP Physics C: Mechanics, AP exams

Unit 7: Oscillations

Unit 7 treats simple harmonic motion as the solution of a differential equation rather than as a set of memorized results. Mass-spring systems, simple pendulums, and physical pendulums are analyzed, and the energy exchange over a cycle is described quantitatively. Recognizing when a restoring force is approximately linear is the key skill.

Position: 7 of 7 units
Topics covered: 4 topics in this guide

What is covered

Topics inside this unit.

These topics describe the scope students most often review for this part of AP Physics C: Mechanics. Treat them as a study checklist rather than an exhaustive list of what a single exam form can sample.

Conditions for simple harmonic motion: A restoring force proportional to displacement, and the differential equation it produces.
Mass-spring systems: Angular frequency, period, and the position, velocity, and acceleration functions over time.
Simple and physical pendulums: The small-angle approximation, and how rotational inertia sets the period of a physical pendulum.
Energy in oscillating systems: Exchange between kinetic and potential energy, and the dependence of total energy on amplitude.

Study tips

How to practice Oscillations.

Section-specific guidance that follows from how this material is usually tested.

Show that a given restoring force is linear in displacement before applying any simple harmonic result.
Practice physical pendulum problems, since they combine rotational inertia with oscillation and appear frequently.
Sketch position, velocity, and acceleration on shared axes and be able to state the phase relationships.

Where this fits

Unit 7: Oscillations is 7 of 7 units in the AP Physics C: Mechanics guide. Use the full guide to balance it against everything else you have to review.

Back to the AP Physics C: Mechanics guide Official exam information

Sample Oscillations questions

Practice questions, not yet expert-reviewed

Original practice questions written to the published framework for this section. No expert has reviewed them, and they are not taken from any official exam. Try each one before you open the answer.

Question 1Foundational
A 0.25 kg block attached to an ideal spring of spring constant 100 N/m oscillates on a frictionless surface. What is the angular frequency of the oscillation?
1. A0.050 rad/s
2. B20 rad/s
3. C25 rad/s
4. D400 rad/s
Show answer
Answer: B. 20 rad/s
The equation of motion gives omega = sqrt(k/m) = sqrt(100/0.25) = sqrt(400) = 20 rad/s. The 400 rad/s choice reports k/m without taking the square root.
Key insight: Angular frequency comes straight from the differential equation, and the period is 2 pi divided by it.
Question 2Foundational
A simple pendulum of length 1.0 m swings with small amplitude where g = 9.8 m/s^2. What is its period of oscillation?
1. A0.50 s
2. B1.0 s
3. C2.0 s
4. D6.3 s
Show answer
Answer: C. 2.0 s
The small-angle period is T = 2(pi)sqrt(L/g) = 2(pi)sqrt(1.0/9.8) = 2.0 s. The 6.3 s choice leaves out the square root factor and reports 2 pi seconds.
Key insight: The small-angle approximation is what makes the pendulum a harmonic oscillator with an amplitude-independent period.
Question 3Intermediate
A block undergoes simple harmonic motion described by x(t) = 0.20 cos(5t), where x is in meters and t is in seconds. What is the maximum speed of the block?
1. A0.040 m/s
2. B0.20 m/s
3. C1.0 m/s
4. D5.0 m/s
Show answer
Answer: C. 1.0 m/s
Differentiating gives v(t) = -1.0 sin(5t), so the maximum speed is (omega)A = (5)(0.20) = 1.0 m/s. The 0.20 m/s choice reports the amplitude, which is a distance rather than a speed.
Key insight: In simple harmonic motion each derivative multiplies the amplitude by another factor of omega.
Question 4Advanced
A uniform rod of mass M and length L hangs from a frictionless pivot at one end and swings with small amplitude as a physical pendulum, with rotational inertia ML^2/3 about the pivot. What is its period of oscillation?
1. A2 pi sqrt(L/g)
2. B2 pi sqrt(L/(2g))
3. C2 pi sqrt(2L/(3g))
4. D2 pi sqrt(3L/(2g))
Show answer
Answer: C. 2 pi sqrt(2L/(3g))
A physical pendulum has T = 2(pi)sqrt(I/(Mgd)) with d the pivot-to-center-of-mass distance, here L/2, so T = 2(pi)sqrt((ML^2/3)/(MgL/2)) = 2(pi)sqrt(2L/(3g)). The 2 pi sqrt(L/g) choice is the simple pendulum result, which treats the rod as a point mass at its end.
Key insight: A swinging rod oscillates faster than a simple pendulum of the same length because its mass is distributed closer to the pivot.
Question 5Intermediate
A 0.50 kg block on a frictionless surface oscillates on a spring of spring constant 200 N/m with an amplitude of 0.10 m. What is the block's speed when it is 0.060 m from the equilibrium position?
1. A0.80 m/s
2. B1.2 m/s
3. C1.6 m/s
4. D2.0 m/s
Show answer
Answer: C. 1.6 m/s
The total energy is (1/2)(200)(0.10)^2 = 1.0 J and the stored potential energy at x = 0.060 m is (1/2)(200)(0.060)^2 = 0.36 J, leaving 0.64 J of kinetic energy, so v = sqrt(2 x 0.64/0.50) = 1.6 m/s. The 2.0 m/s choice is the maximum speed at equilibrium, where the potential energy is zero.
Key insight: Energy conservation gives the speed at any displacement without solving for the time.

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Unit 7: Oscillations