Oscillations
Unit 7 of AP Physics 1, worth 5–8% of the exam. 14 questions below, each with the working. Every answer was checked by a second pass before it was published.
Simple harmonic motion, frequency and period, representing and analysing SHM, energy in simple harmonic oscillators.
How this unit is tested
Start by checking whether a system actually satisfies the condition for simple harmonic motion: the net force (or torque) must be proportional to displacement from equilibrium and point back toward it. This is what makes springs obeying Hooke's Law and pendulums at small angles behave sinusoidally, and it's why problems about non-linear restoring forces are NOT SHM even if they look periodic.
Once you've confirmed SHM, separate the problem into three question types: (1) period/frequency questions, which use T = 2π√(m/k) for springs or T = 2π√(L/g) for pendulums and never depend on amplitude; (2) graph and phase questions, where you track how position, velocity, and acceleration rise and fall together but shifted from each other; and (3) energy questions, where you use conservation of total mechanical energy E = (1/2)kA² to trade kinetic and potential energy at any position x.
For every problem, sketch (even mentally) where the object is: at the equilibrium point, speed is maximum and acceleration is zero; at the extremes (x = ±A), speed is zero and acceleration is maximum. Almost every conceptual trap in this unit comes from mixing up these two locations.
What you have to know
Restoring force (Hooke's Law)
$F = -kx$, where the force on the oscillator is proportional to the displacement x from equilibrium and directed opposite to it (toward equilibrium).
Period–frequency relation
$T = 1/f$, where T is the period (time for one full cycle) and f is the frequency (cycles per second).
Period of a mass-spring system
$T = 2\pi\sqrt{m/k}$, independent of amplitude; depends only on mass m and spring constant k.
Period of a simple pendulum (small angle)
$T = 2\pi\sqrt{L/g}$, valid for angles under about 15°; independent of mass and amplitude.
Energy conservation in SHM
$E_{total} = \tfrac{1}{2}kA^2 = \tfrac{1}{2}kx^2 + \tfrac{1}{2}mv^2$, where the sum of elastic potential and kinetic energy is constant throughout the motion.
Extreme values of speed and acceleration
Maximum speed $v_{max}=A\sqrt{k/m}$ occurs at x = 0; maximum acceleration $a_{max}=kA/m$ occurs at x = ±A.
14 practice questions
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Which condition must be satisfied for a system's motion to be classified as simple harmonic motion?
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Answer. The net force must be proportional to the displacement from equilibrium and directed toward it.
SHM requires a linear restoring force, F = -kx. This is what produces the characteristic sinusoidal motion; forces that vary in other ways (constant, proportional to velocity, etc.) do not produce true SHM. -
A block of mass 0.50 kg is attached to a spring with spring constant 200 N/m. Find the period of oscillation.
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Answer. T ≈ 0.31 s
Use T = 2π√(m/k) = 2π√(0.50/200) = 2π√(0.0025) = 2π(0.05) ≈ 0.31 s. Amplitude is not needed because period doesn't depend on it. -
An oscillator completes one full cycle every 0.25 s. What is its frequency?
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Answer. 4 Hz
Frequency is the reciprocal of period: f = 1/T = 1/0.25 = 4 Hz. -
A simple pendulum's length is quadrupled while g stays the same. What happens to its period?
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Answer. The period doubles.
T = 2π√(L/g), so T is proportional to √L. Quadrupling L multiplies √L by 2, so the period doubles. -
The amplitude of oscillation of a mass on an ideal spring is doubled, with mass and spring constant unchanged. What happens to the period?
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Answer. It stays the same.
Period of an ideal spring-mass system, T = 2π√(m/k), does not depend on amplitude at all — only on m and k. -
A 0.60 kg mass oscillates on a spring with k = 150 N/m and amplitude 0.20 m. Find its maximum speed.
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Answer. ≈3.2 m/s
At equilibrium, all energy is kinetic: (1/2)kA² = (1/2)mv_max². Solving, v_max = A√(k/m) = 0.20√(150/0.60) = 0.20√250 ≈ 0.20(15.8) ≈ 3.2 m/s. -
Which graph is shifted 90° in phase relative to the position-time graph of an SHM oscillator?
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Answer. The velocity-time graph.
Velocity is the rate of change of position; for sinusoidal motion this produces a graph shifted a quarter cycle (90°) from the position graph. Acceleration, by contrast, is 180° out of phase with position (it mirrors position but inverted). -
A force-displacement graph for a spring shows that at x = 0.10 m, the restoring force has magnitude 5.0 N. Find the spring constant.
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Answer. k = 50 N/m
Hooke's Law gives |F| = kx, so k = F/x = 5.0/0.10 = 50 N/m. On a force-displacement graph, k is simply the magnitude of the slope. -
At what point in a spring-mass oscillator's cycle is the magnitude of acceleration greatest?
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Answer. At the maximum displacement (the turning points, x = ±A).
Acceleration is proportional to the net force, which is proportional to displacement (a = -kx/m). It is largest in magnitude where |x| is largest, i.e., at the extremes of motion, and zero at equilibrium. -
A 0.80 kg block oscillates on a spring with k = 40 N/m and amplitude 0.30 m. Find its speed when x = 0.15 m.
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Answer. ≈1.8 m/s
Total energy E = (1/2)kA² = 0.5(40)(0.09) = 1.8 J. Potential energy at x = 0.15 m: (1/2)kx² = 0.5(40)(0.0225) = 0.45 J. Kinetic energy = 1.8 - 0.45 = 1.35 J, so v = √(2(1.35)/0.8) ≈ 1.8 m/s. -
A simple pendulum is taken from Earth (g ≈ 9.8 m/s²) to the Moon (g ≈ 1.6 m/s²) without changing its length. How does its period change?
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Answer. The period becomes longer on the Moon.
T = 2π√(L/g), so T is inversely proportional to √g. Since g is smaller on the Moon, √g decreases, making T larger — the pendulum swings more slowly. -
An oscillator's position is described by x(t) = 0.15 cos((2π/0.40)t) meters. What are its amplitude and period?
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Answer. Amplitude = 0.15 m, period = 0.40 s.
For x(t) = A cos((2π/T)t), the coefficient in front of the cosine is the amplitude, and the coefficient of t inside the cosine equals 2π/T, so T can be read directly from the given expression. -
For a mass on an ideal spring undergoing SHM, what fraction of the total mechanical energy is potential energy when the displacement is exactly half the amplitude (x = A/2)?
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Answer. One-fourth (25%) of the total energy.
PE = (1/2)kx² and total energy E = (1/2)kA². At x = A/2, PE/E = (A/2)²/A² = 1/4. The remaining 3/4 of the energy is kinetic at that point. -
Explain why the period of a simple pendulum does not depend on the mass of the bob.
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Answer. Because both the restoring force (gravity's component along the swing) and the object's inertia are proportional to mass, the mass cancels out of Newton's second law, leaving the period dependent only on length and g.
Applying F = ma to the pendulum's restoring force gives an acceleration equation where mass appears on both sides and cancels, so T = 2π√(L/g) contains no mass term — this mirrors why all objects fall with the same acceleration in free fall.
What people get wrong
- Assuming a larger amplitude makes the period longer or shorter. For ideal SHM (spring or small-angle pendulum), period depends only on m,k or L,g — never on amplitude. Check the formula, not intuition, before answering period questions.
- Thinking speed is maximum at the turning points (x = ±A). Speed is actually zero there and maximum at equilibrium (x = 0); acceleration is the one that's maximum at the extremes. Sketch the motion before assigning max/min values.
- Mixing up which variables affect which period. Mass matters for a spring's period but not for a pendulum's; length matters for a pendulum but has no meaning for a spring. Always match the formula to the physical system.
- Applying the small-angle pendulum formula T = 2π√(L/g) to large swing angles. The formula breaks down above roughly 15°, and the real period becomes amplitude-dependent — don't assume the simple formula always holds.
- Forgetting the negative sign's meaning in F = -kx and getting the restoring force's direction backwards on a force diagram. The force always points toward equilibrium, opposite the displacement, not in the direction of motion.
- Treating total mechanical energy as changing over the cycle. In ideal SHM, E = (1/2)kA² is constant; only the split between kinetic and potential energy changes as the object moves.
Drill this unit until it sticks
These questions come back on a schedule built from what you get wrong, alongside the rest of AP Physics 1. Free, and no account needed to start.