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UNIT 7About 12 min + practice

Oscillations

A restoring interaction turns displacement into repeating motion.

What you’ll learn

  • Identify the conditions for simple harmonic motion.
  • Connect phase, energy, and restoring force.
  • Use spring and pendulum models within their valid ranges.
01

Before you begin

Equilibrium is the position where the net restoring force is zero. Simple harmonic motion requires a restoring acceleration proportional to displacement and opposite in direction. Amplitude is maximum displacement magnitude; period is the time for one full cycle.

Explain these starting ideas in your own words. Revisit them whenever a later step feels unclear.

02

The restoring force points toward equilibrium

Simple harmonic motion occurs when acceleration is proportional to displacement from equilibrium and opposite in direction. For an ideal spring, Hooke’s law supplies this relation. A restoring force need not always be a literal spring; the net force near a stable equilibrium can sometimes be approximated this way.

Equilibrium is where the net force is zero, not necessarily where each individual force is zero. A vertical spring’s equilibrium is displaced by gravity, but small oscillations about that equilibrium can have the same ideal period as a horizontal mass–spring system with the same mass and spring constant.

Fnet = −kx
a = −ω2x

PAUSE & TRY IT

What direction is acceleration when displacement is negative?

Reveal answer

Positive, toward equilibrium, in ideal SHM.

03

Position, velocity, and acceleration are out of step

At maximum displacement, speed is zero and acceleration magnitude is greatest, directed toward equilibrium. At equilibrium, speed is greatest and acceleration is zero in ideal SHM. A zero instantaneous velocity at an endpoint does not mean the object will remain there.

Period is the time for one full cycle, and frequency is cycles per unit time. Angular frequency is 2π times ordinary frequency. Amplitude describes maximum displacement, not total distance traveled in one cycle; that distance is four amplitudes for one-dimensional oscillation.

PAUSE & TRY IT

How far does an oscillator travel in one complete cycle?

Reveal answer

4A for one-dimensional oscillation between −A and +A.

04

Energy trades between kinetic and potential forms

For an ideal undamped spring oscillator, total mechanical energy is ½kA2. Spring potential energy is greatest at the endpoints, while kinetic energy is greatest at equilibrium. At an intermediate displacement, energy conservation determines speed without requiring the elapsed time.

Doubling amplitude quadruples total energy and doubles maximum speed for the same mass and spring. In the ideal linear spring model, the period does not change with amplitude. Real springs can deviate from linear behavior at large displacements.

E = kA2
vmax = A√()
05

Different oscillators have different parameters

An ideal mass–spring oscillator has period 2π√(). A simple pendulum with a point-like bob and light string has approximate period 2π√() at small angular amplitudes. Pendulum mass does not appear in this ideal period, while length does.

The small-angle assumption matters: the restoring torque contains sinθ, approximated by θ in radians only for small angles. Large-amplitude pendulum motion is not described exactly by the simple formula. State the model and its restrictions rather than presenting the formula as universally exact.

A full cycle returns to the same phaseIllustrative SHM: x=0.20 cos(πt). Amplitude is 0.20 m and period is 2 s. Zero displacement is not zero speed.
A full cycle returns to the same phase-0.2-0.100.10.201234 Time (s)Displacement (m)x(t)
Read figure values as text

x(t): 0: 0.2; 0.05: 0.19753766811902757; 0.1: 0.1902113032590307; 0.15: 0.1782013048376736; 0.2: 0.1618033988749895; 0.25: 0.14142135623730953; 0.3: 0.11755705045849463; 0.35: 0.09079809994790937; 0.4: 0.06180339887498949; 0.45: 0.031286893008046185; 0.5: 1.2246467991473533e-17; 0.55: -0.031286893008046206; 0.6: -0.06180339887498947; 0.65: -0.09079809994790934; 0.7: -0.11755705045849461; 0.75: -0.1414213562373095; 0.8: -0.16180339887498948; 0.85: -0.17820130483767357; 0.9: -0.1902113032590307; 0.95: -0.19753766811902754; 1: -0.2; 1.05: -0.19753766811902757; 1.1: -0.1902113032590307; 1.15: -0.17820130483767363; 1.2: -0.1618033988749895; 1.25: -0.14142135623730953; 1.3: -0.11755705045849466; 1.35: -0.09079809994790938; 1.4: -0.061803398874989514; 1.45: -0.031286893008046206; 1.5: -3.6739403974420595e-17; 1.55: 0.031286893008046136; 1.6: 0.06180339887498945; 1.65: 0.09079809994790933; 1.7: 0.11755705045849459; 1.75: 0.14142135623730948; 1.8: 0.16180339887498948; 1.85: 0.17820130483767357; 1.9: 0.1902113032590307; 1.95: 0.19753766811902754; 2: 0.2; 2.05: 0.1975376681190276; 2.1: 0.19021130325903074; 2.15: 0.1782013048376736; 2.2: 0.1618033988749894; 2.25: 0.14142135623730953; 2.3: 0.11755705045849481; 2.35: 0.09079809994790941; 2.4: 0.06180339887498955; 2.45: 0.03128689300804606; 2.5: 6.123233995736766e-17; 2.55: -0.031286893008045935; 2.6: -0.061803398874989424; 2.65: -0.09079809994790915; 2.7: -0.11755705045849459; 2.75: -0.14142135623730934; 2.8: -0.16180339887498946; 2.85: -0.17820130483767363; 2.9: -0.19021130325903068; 2.95: -0.19753766811902757; 3: -0.2; 3.05: -0.1975376681190276; 3.1: -0.19021130325903074; 3.15: -0.1782013048376737; 3.2: -0.16180339887498954; 3.25: -0.14142135623730942; 3.3: -0.1175570504584947; 3.35: -0.09079809994790927; 3.4: -0.06180339887498957; 3.45: -0.03128689300804608; 3.5: -8.572527594031473e-17; 3.55: 0.031286893008045914; 3.6: 0.0618033988749894; 3.65: 0.09079809994790912; 3.7: 0.11755705045849457; 3.75: 0.1414213562373093; 3.8: 0.16180339887498943; 3.85: 0.1782013048376736; 3.9: 0.19021130325903068; 3.95: 0.19753766811902757; 4: 0.2

06

Real oscillations exchange energy with surroundings

Damping reduces mechanical energy over time and generally causes amplitude to decrease. A periodic driving interaction can transfer energy into an oscillator. The response depends on driving frequency and damping; a large response can occur near resonance.

To measure a period, time several cycles and divide by the number of complete cycles, reducing the relative effect of start/stop reaction time. Keep amplitude within the intended model range and measure pendulum length to the bob’s center of mass, not merely to the knot.

Energy exchanges during a spring oscillationIllustrative model, not collected experimental data. For k=20 N/m and amplitude 0.5 m, total energy is 2.5 J. Kinetic energy is greatest at equilibrium and zero at the turning points.
Energy exchanges during a spring oscillation0123-0.5-0.2500.250.5 Displacement (m)Energy (J)Spring potentialKinetic
Read figure values as text

Spring potential: -0.5: 2.5; -0.4791666666666667: 2.2960069444444446; -0.4583333333333333: 2.100694444444444; -0.4375: 1.9140625; -0.4166666666666667: 1.7361111111111114; -0.3958333333333333: 1.5668402777777777; -0.375: 1.40625; -0.35416666666666663: 1.2543402777777775; -0.33333333333333337: 1.1111111111111114; -0.3125: 0.9765625; -0.29166666666666663: 0.8506944444444442; -0.27083333333333337: 0.7335069444444448; -0.25: 0.625; -0.22916666666666669: 0.5251736111111113; -0.20833333333333331: 0.4340277777777777; -0.1875: 0.3515625; -0.16666666666666669: 0.27777777777777785; -0.14583333333333331: 0.21267361111111105; -0.125: 0.15625; -0.10416666666666669: 0.10850694444444449; -0.08333333333333331: 0.06944444444444442; -0.0625: 0.0390625; -0.041666666666666685: 0.017361111111111126; -0.020833333333333315: 0.00434027777777777; 0: 0; 0.02083333333333337: 0.004340277777777794; 0.04166666666666663: 0.01736111111111108; 0.0625: 0.0390625; 0.08333333333333337: 0.0694444444444445; 0.10416666666666663: 0.10850694444444436; 0.125: 0.15625; 0.14583333333333337: 0.21267361111111122; 0.16666666666666663: 0.2777777777777777; 0.1875: 0.3515625; 0.20833333333333337: 0.43402777777777796; 0.22916666666666663: 0.5251736111111109; 0.25: 0.625; 0.27083333333333337: 0.7335069444444448; 0.29166666666666663: 0.8506944444444442; 0.3125: 0.9765625; 0.33333333333333337: 1.1111111111111114; 0.35416666666666663: 1.2543402777777775; 0.375: 1.40625; 0.39583333333333337: 1.5668402777777781; 0.41666666666666663: 1.7361111111111107; 0.4375: 1.9140625; 0.45833333333333337: 2.100694444444445; 0.47916666666666663: 2.296006944444444; 0.5: 2.5 • Kinetic: -0.5: 0; -0.4791666666666667: 0.20399305555555536; -0.4583333333333333: 0.3993055555555558; -0.4375: 0.5859375; -0.4166666666666667: 0.7638888888888886; -0.3958333333333333: 0.9331597222222223; -0.375: 1.09375; -0.35416666666666663: 1.2456597222222225; -0.33333333333333337: 1.3888888888888886; -0.3125: 1.5234375; -0.29166666666666663: 1.6493055555555558; -0.27083333333333337: 1.7664930555555554; -0.25: 1.875; -0.22916666666666669: 1.9748263888888888; -0.20833333333333331: 2.0659722222222223; -0.1875: 2.1484375; -0.16666666666666669: 2.2222222222222223; -0.14583333333333331: 2.287326388888889; -0.125: 2.34375; -0.10416666666666669: 2.3914930555555554; -0.08333333333333331: 2.4305555555555554; -0.0625: 2.4609375; -0.041666666666666685: 2.482638888888889; -0.020833333333333315: 2.4956597222222223; 0: 2.5; 0.02083333333333337: 2.4956597222222223; 0.04166666666666663: 2.482638888888889; 0.0625: 2.4609375; 0.08333333333333337: 2.4305555555555554; 0.10416666666666663: 2.391493055555556; 0.125: 2.34375; 0.14583333333333337: 2.287326388888889; 0.16666666666666663: 2.2222222222222223; 0.1875: 2.1484375; 0.20833333333333337: 2.065972222222222; 0.22916666666666663: 1.974826388888889; 0.25: 1.875; 0.27083333333333337: 1.7664930555555554; 0.29166666666666663: 1.6493055555555558; 0.3125: 1.5234375; 0.33333333333333337: 1.3888888888888886; 0.35416666666666663: 1.2456597222222225; 0.375: 1.09375; 0.39583333333333337: 0.9331597222222219; 0.41666666666666663: 0.7638888888888893; 0.4375: 0.5859375; 0.45833333333333337: 0.3993055555555549; 0.47916666666666663: 0.2039930555555558; 0.5: 0

PAUSE & TRY IT

Why time many pendulum cycles rather than one?

Reveal answer

It reduces the relative contribution of timing reaction error to the estimated period.

07

Read one cycle through position, velocity, and force

At maximum displacement, speed is zero and restoring-force magnitude is largest. At equilibrium, speed is greatest while the restoring force is zero in the ideal spring model. Acceleration points toward equilibrium, so it is opposite in sign to displacement. These facts let you reason about phase relationships without memorizing every sine and cosine expression.

A full cycle returns the object to the same position and direction of motion. Passing equilibrium twice does not necessarily mean two periods have elapsed. Position, velocity, and acceleration graphs are shifted relative to one another, and their maxima do not all occur at the same time. Use slopes and the restoring-force relation to connect them.

PAUSE & TRY IT

Where is acceleration magnitude largest in ideal spring motion?

Reveal answer

At the endpoints, where displacement magnitude is largest.

08

Derive parameter effects from the period model

For a mass on an ideal spring, T=2π√(). Multiplying mass by four doubles period; multiplying spring constant by four halves period. Amplitude does not change period in the ideal linear-spring model. If a real spring leaves its linear regime, that idealization can fail.

For a simple pendulum at small angles, T=2π√(). Mass does not appear, but length and gravitational field strength do. A larger initial angle can invalidate the small-angle approximation. When comparing experiments, identify which quantities are held fixed and whether the model assumptions remain valid across the range.

PAUSE & TRY IT

Does doubling pendulum mass double its small-angle period?

Reveal answer

No. Mass does not appear in the ideal small-angle period formula.

09

Energy and damping explain real motion

For a horizontal ideal spring oscillator, total mechanical energy is . At a displacement x, spring potential energy is and the remaining mechanical energy is kinetic. Speed therefore depends on position, not just on whether the object is moving left or right. Equal displacement magnitudes give equal speed magnitudes in the ideal model.

Damping transfers mechanical energy to surroundings and reduces amplitude over time. A driven oscillator can receive energy repeatedly, and resonance can produce large responses near an appropriate driving frequency. A decreasing amplitude is evidence of energy transfer, not proof that the spring constant is changing. Distinguish the ideal model from the effect being investigated.

10

Identify a restoring interaction before naming simple harmonic motion

Simple harmonic motion requires a restoring force proportional to displacement from equilibrium with the opposite sign. For an ideal spring, F=−kx gives acceleration a=−()x. Equilibrium is where the net force is zero, not necessarily where every individual force vanishes.

At maximum displacement, speed is zero but acceleration magnitude is greatest. At equilibrium, speed is greatest and acceleration is zero for the ideal oscillator. These phase relationships explain why position, velocity and acceleration graphs are shifted relative to one another.

For a vertical spring, gravity changes the equilibrium extension. Measuring displacement from that new equilibrium produces the same ideal oscillation period as the corresponding horizontal spring of the same m and k. Do not use total spring extension and displacement from equilibrium interchangeably.

PAUSE & TRY IT

Where is an ideal spring oscillator accelerating most strongly?

Reveal answer

At the turning points, where |x| is largest. Speed is zero there, but |a|=()|x| is maximal.

11

Use period relationships within their approximations

A spring–mass period is 2π√() under the ideal model. Increasing mass increases period; increasing stiffness decreases it. The ideal period is independent of amplitude, but very large deformations can violate the linear spring assumption.

A simple pendulum has period approximately 2π√() for small angular amplitudes with the usual point-mass and light-string assumptions. Mass cancels because it appears in both gravitational force and inertia. The small-angle approximation is part of the model; it is not justified solely because a diagram draws a short arc.

In an experiment, graphing T2 against m for a spring or T2 against L for a simple pendulum can linearize the theoretical relation. Identify what the slope represents and whether an intercept might reveal an unmodeled effective mass or systematic offset. A line fit is useful only when the model and axes are appropriate.

12

Track energy and interpret real departures

Ideal oscillation exchanges kinetic and potential energy while total mechanical energy stays constant. For a spring oscillator, total energy is ½kA2, so doubling amplitude quadruples energy. The object does not move at constant speed between turning points.

Real damping transfers mechanical energy to other forms and generally decreases amplitude over time. An observed decay is not proof that total energy conservation fails. Identify the interaction responsible and the larger system receiving the energy.

A graph’s period is the time between equivalent phase points, such as successive positive maxima, not merely adjacent zero crossings. Frequency is the reciprocal of period. Angular frequency is 2π times frequency; confusing these quantities introduces a factor of 2π into calculations.

13

Identify the restoring relationship before using SHM formulas

Simple harmonic motion occurs when acceleration is proportional to displacement from equilibrium and opposite in direction: a=−ω2x. A restoring tendency alone does not guarantee exact SHM; the force-displacement relationship must have the appropriate linear form over the range considered.

For a mass–spring system, ω=√() and T=2π√(). A vertical spring’s equilibrium shifts because of gravity, but displacement measured from that equilibrium follows the same ideal period. Do not measure the oscillation coordinate from the spring’s unstretched length unless the force equation accounts for gravity explicitly.

A simple pendulum has approximately harmonic motion for small angles, with T≈2π√(). The approximation weakens at larger amplitudes. The mass does not appear in this ideal period, but damping, finite amplitude, and other nonideal effects can alter observed behavior.

14

Connect phase, energy, and measurement

At the turning points, speed is zero while acceleration magnitude is largest toward equilibrium. At equilibrium, speed is greatest while acceleration is zero in the ideal linear model. Velocity and acceleration do not reach their extrema at the same times.

For x=A cos(ωt+φ), the phase φ encodes the initial state. A graph starting at maximum displacement differs from one crossing equilibrium, even with the same amplitude and period. Use initial position and direction of motion together to choose phase.

To measure a period, time several cycles and divide by the number of cycles to reduce relative timing uncertainty. Count complete cycles consistently. Vary one parameter while controlling others to test a proportionality such as T2∝m for a fixed spring.

PAUSE & TRY IT

Where is acceleration zero in ideal mass–spring SHM?

Reveal answer

At equilibrium, where displacement from equilibrium is zero. Speed is greatest there.

FROM IDEA TO APPLICATION

Worked examples

EXAMPLE 1

A parameter comparison

The mass on an ideal spring is increased by a factor of four while k remains fixed. How does period change?

Reveal worked solution
  1. T is proportional to √m.
  2. √4 = 2.
Result & interpretation

The period doubles. The conclusion does not require a change in amplitude.

EXAMPLE 2

Speed halfway to the endpoint

An ideal spring oscillator is at x = . What fraction of total energy is kinetic?

Reveal worked solution
  1. Potential fraction = = .
  2. Kinetic fraction = 1 − .
Result & interpretation

of the total energy is kinetic, and speed is of maximum speed.

EXAMPLE 3

Speed at an intermediate displacement

An ideal spring oscillator has k=100 , m=1.0 kg, and amplitude 0.20 m. Find speed when x=0.12 m.

Reveal worked solution
  1. Total energy=()(100)(0.20)2=2.0 J.
  2. Spring energy at x=0.12 m is 0.72 J.
  3. Kinetic energy is 1.28 J, so v2=2.
Result & interpretation

Speed is 1.6 ; velocity can be positive or negative depending on direction.

EXAMPLE 4

Predict a period change without calculating every constant

An ideal spring oscillator’s mass is multiplied by four while k is unchanged. Its original period is 0.60 s. Find the new period.

Reveal worked solution
  1. Period is proportional to √m at fixed k.
  2. Multiplying mass by four multiplies period by √4=2.
  3. The new period is 1.20 s.
Result & interpretation

The oscillator takes twice as long per cycle.

EXAMPLE 5

Infer a spring constant from a period

A 0.5 kg mass oscillates with period 1 s on an ideal spring.

Reveal worked solution
  1. T2=.
  2. Rearrange k=.
Result & interpretation

k=2π2 , about 19.7 .

EXAMPLE 6

Change pendulum length

A small-angle pendulum’s length becomes nine times larger at the same g.

Reveal worked solution
  1. T is proportional to √L.
  2. √9=3.
Result & interpretation

Its period triples.

MAKE THE DISTINCTION

Common mistakes, clearer reasoning

The trapAt equilibrium the oscillator stops.

The better explanationIts acceleration is zero there, but its speed is greatest in ideal SHM.

The trapThe simple pendulum period formula is exact at every amplitude.

The better explanationIt uses a small-angle approximation.

RETRIEVE BEFORE YOU REVEAL

Practice checkpoints

Revisit the quick checks from this guide without looking back. Explain why, then reveal the answer.

1. What direction is acceleration when displacement is negative?

Reveal answer

Positive, toward equilibrium, in ideal SHM.

2. How far does an oscillator travel in one complete cycle?

Reveal answer

4A for one-dimensional oscillation between −A and +A.

3. Why time many pendulum cycles rather than one?

Reveal answer

It reduces the relative contribution of timing reaction error to the estimated period.

4. Where is acceleration magnitude largest in ideal spring motion?

Reveal answer

At the endpoints, where displacement magnitude is largest.

5. Does doubling pendulum mass double its small-angle period?

Reveal answer

No. Mass does not appear in the ideal small-angle period formula.

6. Where is an ideal spring oscillator accelerating most strongly?

Reveal answer

At the turning points, where |x| is largest. Speed is zero there, but |a|=()|x| is maximal.

7. Where is acceleration zero in ideal mass–spring SHM?

Reveal answer

At equilibrium, where displacement from equilibrium is zero. Speed is greatest there.

Key language

Amplitude
Maximum displacement from equilibrium.
Period
Time for one complete cycle.
Simple harmonic motion
Motion with acceleration proportional and opposite to displacement.
Damping
Energy transfer out of oscillatory mechanical motion.
Connect it to the course

Oscillations combine Newton’s laws, energy conservation, and the geometry of stable equilibrium.

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Written for ScienceHub · Original instructional material. Course framework reference ↗. These notes are independently authored and are not College Board materials. External photographs retain their credited licenses.

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