Oscillations: AP Physics 1 Unit 7

Why this unit matters

Oscillations appear in clocks, instruments, sensors, and many natural systems where repeated motion carries information.

What you will learn

Describe simple harmonic motion using displacement, velocity, acceleration, and phase relationships.
Relate period and frequency to mass-spring and pendulum system parameters.
Track energy exchange between kinetic and potential forms during oscillation.

Understand the core ideas

Oscillatory motion repeats around an equilibrium point because a restoring influence points back toward equilibrium after displacement. In ideal simple harmonic motion, acceleration is proportional to displacement and opposite in sign, which leads to smooth periodic reversal at turning points. Period T is time per cycle, frequency f is cycles per second, and they are related by f = 1/T.

For a mass-spring system, period is T = 2pi $\sqrt{m/k}$ , so increasing mass increases period while increasing spring constant decreases period. For a small-angle pendulum, period depends mainly on length and gravitational field strength, not bob mass. During one cycle, energy shifts between kinetic and potential forms while total mechanical energy remains constant in the ideal undamped model.

Representation links are crucial: displacement-time graphs show phase, velocity is largest at equilibrium, and acceleration magnitude is largest at maximum displacement. Using these checkpoints helps you infer motion state even when equations are not provided directly.

Key terms

amplitude: Maximum displacement from equilibrium in an oscillation.
period: Time required for one complete cycle of motion.
frequency: Number of cycles per second, equal to 1/period.
restoring force: Force that points toward equilibrium and drives oscillatory return motion.

Mass-spring period and frequency

A 0.50 kg mass oscillates on a spring with k = 8.0 N/m on a low-friction surface.

Use the period formula for a mass-spring oscillator: T = 2pi $\sqrt{m/k}$ .
Substitute values: T = 2pi $\sqrt{0.50/8.0}$ = 2pi $\sqrt{0.0625}$ .
Evaluate the square root: $\sqrt{0.0625}$ = 0.25, so T = 2pi(0.25) = 0.5pi s.
Convert to decimal: T is about 1.57 s, then compute frequency f = 1/T about 0.64 Hz.
Check behavior: at maximum displacement speed is zero, while at equilibrium speed is maximum.

Result: The oscillator has period about 1.57 s and frequency about 0.64 Hz.

A common misconception

Claim: Acceleration is zero at turning points because velocity is zero there.

Correction: At turning points, restoring force magnitude is greatest, so acceleration magnitude is also greatest even though velocity is zero.

Lessons in this unit

Periodic motion basicsIdentify amplitude, period, and frequency from representations.
Mass-spring systemsUse restoring-force and period relationships for horizontal and vertical setups.
Pendulum modelsApply small-angle approximations and discuss model limits.
Energy in oscillationsExplain how total energy remains constant in an ideal undamped oscillator.

Study task

For a spring-mass system with known k and m, compute period and sketch energy versus time over one cycle.

Unit checkpoint

A mass-spring oscillator has period T = 2.0 s. What is its frequency?

Frequency f = 1/T = 1/2.0 = 0.50 Hz.

Oscillations