Review AP Physics 1 Unit 7 to build a complete picture of oscillating systems, from the restoring force that defines simple harmonic motion to the energy exchanges that keep a spring or pendulum moving. This unit connects force, kinematics, and energy conservation in one unified framework.
Use the topic guides, practice questions, and FRQ practice available for this unit to work through SHM problems from multiple angles.
Oscillations appear whenever a restoring force pulls an object back toward equilibrium. Unit 7 formalizes that idea into simple harmonic motion, a model that applies to springs, pendulums, and many real systems. The unit builds from defining SHM, to calculating how fast a system oscillates, to representing the motion graphically, to tracking energy throughout the cycle.
Simple harmonic motion occurs when a restoring force is proportional to displacement from equilibrium. The key equations are ma = -kdelta_x for the force condition, T = 2pisqrt(m/k) for a spring, T = 2pisqrt(L/g) for a pendulum, and E_total = (1/2)kA^2 for total mechanical energy. Kinetic energy is maximum at equilibrium; potential energy is maximum at the turning points.
SHM requires a linear restoring force: the net force on the object must point back toward equilibrium and be proportional to displacement. The equation ma_x = -kdelta_x captures this. A pendulum with small angular displacement qualifies because the restoring torque is proportional to the angular displacement.
For a spring-mass system, T = 2pisqrt(m/k). For a simple pendulum, T = 2pisqrt(L/g). Neither formula includes amplitude, which means doubling the amplitude of oscillation does not change how long each cycle takes. This is a frequently tested and counterintuitive result.
Total mechanical energy in SHM is constant and equals (1/2)kA^2 for a spring system. At the turning points (x = plus or minus A), all energy is potential and kinetic energy is zero. At equilibrium (x = 0), all energy is kinetic and speed is maximum. Increasing amplitude increases total energy by a factor of A squared.
The same SHM model describes both spring-mass oscillators and small-angle pendulums. In both cases a restoring force or torque is proportional to displacement, the period is amplitude-independent, and energy continuously converts between kinetic and potential forms while the total stays constant. Recognizing this shared structure lets you transfer reasoning from one system to the other on any problem.
SHM is defined by a linear restoring force: ma_x = -kdelta_x. The force and acceleration are always directed toward equilibrium and are proportional to displacement. Pendulums with small angular displacement qualify under the same logic.
T = 1/f. For a spring, T = 2pisqrt(m/k). For a pendulum, T = 2pisqrt(L/g). Amplitude does not appear in either formula, so changing amplitude does not change how long each oscillation takes.
Displacement, velocity, and acceleration all vary sinusoidally with the same period but different phases. Velocity is zero at turning points and maximum at equilibrium. Acceleration is always opposite to displacement. Amplitude changes the vertical scale of graphs but not the period.
Total mechanical energy is constant at (1/2)kA^2. Energy shifts between kinetic and potential as the object oscillates. Kinetic energy peaks at equilibrium; potential energy peaks at the turning points. Total energy scales with the square of amplitude.
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Review Defining Simple Harmonic Motion with attention to how the concept appears in AP-style source and evidence questions.
Review Frequency and Period of SHM with attention to how the concept appears in AP-style source and evidence questions.
Review Energy of Simple Harmonic Oscillators with attention to how the concept appears in AP-style source and evidence questions.
SHM is a special case of periodic motion defined by its force law. The net force on the object must be a restoring force: it points opposite to the displacement from equilibrium and its magnitude is proportional to that displacement. The governing equation is ma_x = -kdelta_x, where k is the spring constant and delta_x is displacement from equilibrium. Because acceleration is proportional to displacement and oppositely directed, the object accelerates most strongly at the turning points and has zero acceleration at equilibrium. A simple pendulum with small angular displacement fits this model because the restoring torque is proportional to the angular displacement, making the angular acceleration proportional to the angle.
A block on a spring is pulled 0.10 m from equilibrium and released. At what position is the net force on the block greatest in magnitude, and at what position is it zero?
| System | Restoring quantity | Proportionality condition |
|---|---|---|
| Spring-mass | Force F = -k*delta_x | Force proportional to linear displacement |
| Simple pendulum (small angle) | Torque tau proportional to theta | Torque proportional to angular displacement |
Period T and frequency f are inverses: T = 1/f. For a spring-mass system, T_s = 2pisqrt(m/k). For a simple pendulum with small angular displacement, T_p = 2pisqrt(L/g). Neither formula contains amplitude, confirming that amplitude does not affect how long each oscillation takes. To predict how a change in the system affects the period, isolate the variable that changed: doubling mass increases T_s by a factor of sqrt(2); quadrupling spring constant cuts T_s in half; doubling pendulum length increases T_p by sqrt(2); changing amplitude has no effect on either period.
A pendulum has period T on Earth. If you take it to a planet where g is four times larger, what is the new period in terms of T?
| System | Period formula | Variables that change T | Variables that do NOT change T |
|---|---|---|---|
| Spring-mass | 2*pi*sqrt(m/k) | Mass m, spring constant k | Amplitude A |
| Simple pendulum | 2*pi*sqrt(L/g) | Length L, gravitational acceleration g | Bob mass, amplitude A |
Displacement in SHM follows x = Acos(2pift) or x = Asin(2pift) depending on initial conditions. Velocity and acceleration are also sinusoidal but shifted in phase. Velocity is zero at the turning points (x = plus or minus A) and maximum at equilibrium (x = 0). Acceleration is maximum in magnitude at the turning points and zero at equilibrium, always pointing opposite to displacement. On a graph, the x-t, v-t, and a-t curves all have the same period but peak at different times. Changing amplitude stretches the curves vertically but does not change the period or the horizontal spacing of the peaks and zeros.
An object in SHM has displacement x = 0.05cos(4pi*t) meters. At t = 0, what are the displacement, velocity direction, and acceleration direction?
| Quantity | At x = +A (turning point) | At x = 0 (equilibrium) |
|---|---|---|
| Displacement | Maximum positive | Zero |
| Speed | Zero | Maximum |
| Acceleration magnitude | Maximum | Zero |
| Restoring force magnitude | Maximum | Zero |
The total mechanical energy of an SHM system is constant: E_total = U + K. For a spring-mass system, E_total = (1/2)kA^2. At the turning points, all energy is stored as spring potential energy U = (1/2)kx^2 and kinetic energy is zero. At equilibrium, all energy is kinetic and speed is at its maximum, v_max = A*sqrt(k/m). Because E_total scales with A squared, doubling the amplitude quadruples the total energy. This energy framework lets you find the speed of the object at any position using (1/2)kA^2 = (1/2)mv^2 + (1/2)kx^2.
A spring with k = 200 N/m oscillates with amplitude 0.04 m. What is the total mechanical energy, and what is the speed of the mass at x = 0.02 m if the mass is 0.5 kg?
| Position | Kinetic energy | Potential energy | Total energy |
|---|---|---|---|
| x = A (turning point) | Zero | Maximum = (1/2)*k*A^2 | (1/2)*k*A^2 |
| x = 0 (equilibrium) | Maximum = (1/2)*k*A^2 | Zero | (1/2)*k*A^2 |
| x between 0 and A | Partial | Partial | (1/2)*k*A^2 |
Try stimulus-based AP practice questions and written prompts after you review the notes.
A simple pendulum oscillates with a small amplitude. The graph shows the horizontal velocity of the pendulum bob as a function of time .
Which of the following graphs could represent the horizontal acceleration of the pendulum bob as a function of time ?
A block attached to a horizontal spring oscillates on a frictionless surface. The graph shows the block's position as a function of time .
Which of the following graphs could represent the kinetic energy of the block as a function of time ?
A timed, AP-style test built only from unit 7 questions — multiple choice plus free response, scored with AI feedback. Use the standard format or build your own.
| Term | Definition |
|---|---|
| displacement from equilibrium | The distance and direction of an object's position relative to its equilibrium position; in SHM, the restoring force is proportional to this quantity. |
| resonance | The condition in which a driven oscillator vibrates at maximum amplitude when the driving frequency matches a natural resonant frequency of the system. |
Neither T_s = 2pisqrt(m/k) nor T_p = 2pisqrt(L/g) contains amplitude. A larger oscillation takes the same time per cycle as a smaller one. Amplitude does affect total energy and maximum speed, but not period or frequency.
Velocity is maximum at equilibrium (x = 0), not at the turning points. Acceleration is maximum at the turning points (x = plus or minus A), not at equilibrium. These are out of phase by a quarter cycle, and mixing them up leads to wrong answers on graph and scenario questions.
T_p = 2pisqrt(L/g) is valid only for small angular displacements where sin(theta) is approximately equal to theta. For large angles, the motion is no longer simple harmonic and this formula does not apply.
The formula T_p = 2pisqrt(L/g) contains no mass term. Changing the mass of the pendulum bob does not change the period. Students often assume heavier bobs swing more slowly.
The total energy is (1/2)kA^2, where A is the amplitude, not the current displacement x. At an arbitrary position x, the potential energy is (1/2)kx^2 and the kinetic energy makes up the rest. Setting total energy equal to (1/2)kx^2 at a non-turning-point position is a common algebra error.
A common task presents a spring-mass or pendulum system and asks how the period, frequency, or total energy changes when one variable is altered. You need to apply T_s = 2pisqrt(m/k) or T_p = 2pisqrt(L/g) algebraically, explain why amplitude does not appear, and use E_total = (1/2)kA^2 to reason about energy changes. Justifying your answer with the relevant equation is expected.
Multi-part problems often give one representation of SHM, such as an x-t graph or a position equation, and ask you to derive or sketch the corresponding v-t and a-t graphs, identify specific values at labeled points, or describe the motion in words. Knowing the phase relationships between displacement, velocity, and acceleration and being able to identify turning points and equilibrium crossings from any representation is a core skill.
Problems frequently ask for the speed of an oscillating object at a position other than equilibrium or a turning point. The approach is to set (1/2)kA^2 equal to (1/2)mv^2 + (1/2)kx^2 and solve. Variations include changing the amplitude and asking for the new maximum speed, or describing energy bar charts at different points in the cycle and asking for qualitative comparisons.
Use this list to confirm you can handle every major skill in the Oscillations unit before exam day.
Given a force equation or scenario, confirm whether the restoring force is proportional to displacement from equilibrium and directed opposite to it, satisfying ma_x = -kdelta_x.
Apply T_s = 2pisqrt(m/k) for a spring-mass oscillator and T_p = 2pisqrt(L/g) for a simple pendulum. Predict how changing m, k, L, or g affects the period, and confirm that amplitude changes have no effect.
Given an x-t graph, identify where velocity and acceleration are zero, maximum, or minimum. Sketch corresponding v-t and a-t graphs with correct phase relationships and the same period.
Use E_total = (1/2)kA^2 = (1/2)mv^2 + (1/2)kx^2 to find speed at any position. Identify where kinetic and potential energies are maximum or zero, and explain how changing amplitude changes total energy.
Articulate in words and with equations why the period of a spring-mass system or small-angle pendulum does not depend on amplitude, and why total energy does depend on amplitude.
Start with Topic 7.1. Read the topic guide and practice identifying whether a described force qualifies as a restoring force. Write out ma_x = -kdelta_x and explain in your own words why acceleration is largest at the turning points and zero at equilibrium. Apply the same logic to the pendulum case.
Work through Topic 7.2 by solving problems that ask you to calculate T or f for spring-mass and pendulum systems, then predict how the period changes when one variable is doubled or halved. Confirm for yourself that amplitude never appears in either period formula.
For Topic 7.3, practice sketching x-t, v-t, and a-t graphs for the same oscillator and labeling the turning points, equilibrium crossings, and phase relationships. Given one graph, derive the other two. Use the equations x = Acos(2pift) to check your sketches numerically.
Work through Topic 7.4 by applying E_total = (1/2)kA^2 = (1/2)mv^2 + (1/2)kx^2 to find speed at intermediate positions. Practice problems where amplitude changes and you must find the new total energy. Connect energy graphs to position graphs by identifying where KE and PE peak.
Use the 25+ practice questions and FRQ practice available for this unit to work through multi-part problems that combine force, period, graphs, and energy in a single scenario. Use the AP score calculator to estimate your estimated score range and identify which topic areas need more attention.
Open the individual guides for Unit 7 when you want a closer review of one topic.
Use AP-style practice after you review the notes so you can check what you understand.
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get printablesAP Physics 1 Unit 7 covers four topics focused on oscillations and simple harmonic motion: 7.1 Defining Simple Harmonic Motion, 7.2 Frequency and Period of SHM, 7.3 Representing and Analyzing SHM, and 7.4 Energy of Simple Harmonic Oscillators. You'll apply these ideas to spring-object systems and pendulums. The unit connects force, energy, and periodic motion in a way that shows up across many real-world applications. Head to Unit 7 for topic-by-topic practice.
Unit 7 makes up 5-8% of the AP Physics 1 exam. That weight covers oscillations and simple harmonic motion, including how energy transforms in spring-object systems and pendulums, how to calculate frequency and period, and how to represent SHM graphically and mathematically. It's a smaller unit by topic count (4 topics), but the energy and SHM concepts it introduces connect directly to other units, so a solid understanding pays off on exam day.
The AP Physics 1 Unit 7 progress check includes both MCQ and FRQ parts drawn from all four unit topics: defining simple harmonic motion, frequency and period, representing and analyzing SHM, and the energy of simple harmonic oscillators. MCQ questions typically ask you to interpret graphs, compare systems, or calculate period and frequency. FRQ questions often ask you to explain energy transformations or justify the motion of a spring or pendulum using SHM principles. Working through the progress check is one of the best ways to spot gaps before the real exam. Find matched practice at Unit 7.
AP Physics 1 Unit 7 FRQs most often focus on energy in simple harmonic oscillators and representing or analyzing SHM, so those are the two areas to prioritize. Typical question types ask you to sketch or interpret position-time and energy graphs, explain why a restoring force produces SHM, or compare how changing mass or spring constant affects period. To practice effectively, write out full justifications, not just numerical answers. College Board rewards clear reasoning. You can find FRQ-style practice questions at Unit 7.
The best place to find AP Physics 1 Unit 7 practice questions, including multiple-choice and practice test style problems, is Unit 7. That page organizes MCQ and FRQ practice around all four topics: defining SHM, frequency and period, representing and analyzing SHM, and energy of simple harmonic oscillators. For a practice test experience, work through questions from each topic in one sitting and time yourself. Mixing MCQ and FRQ in the same session mirrors what the real exam feels like.
Start with the concept of energy in simple harmonic motion, since it ties together nearly every other idea in the unit. Once you understand how kinetic and potential energy trade off in a spring-object oscillator or pendulum, the rest of the unit, including frequency, period, and SHM graphs, clicks into place much faster. Here's a practical study sequence: 1. **Define SHM** (Topic 7.1): Make sure you can explain what a restoring force is and why it produces oscillations. 2. **Frequency and period** (Topic 7.2): Practice deriving and applying the period formulas for springs and pendulums. 3. **Graphs and representations** (Topic 7.3): Sketch position, velocity, and acceleration vs. time graphs from scratch until it feels automatic. 4. **Energy** (Topic 7.4): Work problems where you track energy at different points in the oscillation cycle. After each topic, do a short set of MCQ to check your understanding before moving on. Find topic-aligned practice at Unit 7.