Simple Harmonic Motion — AP Physics 1 Study Guide

For: AP Physics 1 candidates sitting AP Physics 1.

Covers: Restoring force and Hooke's law, period and frequency of springs and pendulums, energy in SHM, graphical and equation-based analysis — AP Physics 1 Unit 6 (CED 2024-25).

You should already know: Hooke's law and springs (Unit 2), conservation of energy (Unit 4).

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the AP Physics 1 style for educational use. They are not reproductions of past College Board papers.

1. What Makes Motion "Simple Harmonic"?

Simple harmonic motion (SHM) is any motion where the restoring force is proportional to displacement and points back towards equilibrium: $F=-kx$. This minus sign is the defining feature — push it right and the force pushes left, push it left and the force pushes right.

Two canonical examples that satisfy $F=-kx$ exactly: an ideal spring obeying Hooke's law, and a pendulum at small angles (small-angle approximation $\sin \theta \approx \theta$).

Result: position oscillates sinusoidally over time: $$x(t)=A\cos (\omega t+\varphi )$$ with amplitude $A$, angular frequency $\omega$, and phase constant $\varphi$.

2. Period and frequency

Period $T$ = time for one full oscillation. Frequency $f=1/T$ (Hz). Angular frequency $\omega =2\pi f=2\pi /T$.

Mass on spring: $T=2\pi \sqrt{m/k}$. Period depends on mass and spring constant only — not amplitude.

Simple pendulum (length $L$, small angles): $T=2\pi \sqrt{L/g}$. Depends on length and gravity only — not mass or amplitude.

For both, increasing the "stiffness" ($k$ for springs, $g/L$ for pendulums) shortens the period.

3. Velocity and acceleration

Differentiating $x(t)=A\cos (\omega t+\varphi )$: $$v(t)=-A\omega \sin (\omega t+\varphi ),a(t)=-A{\omega }^2\cos (\omega t+\varphi )=-{\omega }^{2}x$$

Maxima: v_\max = A\omega at equilibrium ($x=0$); a_\max = A\omega^2 at the turning points ($x=\pm A$).

Energy in SHM: $E_{\text{total}}=\frac{1}{2}k{A}^2$ (constant). Splits between KE (max at $x=0$) and PE (max at $x=\pm A$). At any position: $$\frac{1}{2}m{v}^2+\frac{1}{2}k{x}^2=\frac{1}{2}k{A}^2$$

4. Graphical features

The position-time graph of SHM is a sinusoid. Key features to identify on FRQ:

• Amplitude = max displacement from equilibrium.
• Period = horizontal distance between corresponding points (peak to peak, zero crossing in same direction to next zero crossing in same direction).
• At extrema: velocity is zero, acceleration is maximum opposite to displacement.
• At equilibrium: velocity maximum, acceleration zero.

Velocity-time and acceleration-time graphs are also sinusoidal but phase-shifted by π/2 and π respectively from position.

5. Worked Example

A 0.50 kg mass is attached to a horizontal spring with $k=200$ N/m and pulled 0.10 m from equilibrium, then released. (a) Calculate period $T$ and angular frequency $\omega$. (b) Find the maximum speed of the mass. (c) Find the speed when the mass is 0.05 m from equilibrium.

Solution.

(a) $T=2\pi \sqrt{m/k}=2\pi \sqrt{0.50/200}=2\pi \cdot 0.05=0.314$ s. $\omega =2\pi /T=20$ rad/s.

(b) v_\max = A\omega = 0.10 \cdot 20 = 2.0 m/s.

(c) Energy conservation: $\frac{1}{2}k{A}^2=\frac{1}{2}k{x}^2+\frac{1}{2}m{v}^2$. $\frac{1}{2}(200)(0.10{)}^2=\frac{1}{2}(200)(0.05{)}^2+\frac{1}{2}(0.50)v^2$ $1.0=0.25+0.25v^2$ $v^2=3.0$, $v=1.73$ m/s.

6. Common Pitfalls

• Pendulum period and mass: pendulum period is independent of mass. AP graders catch this often — don't add mass into the formula.
• Small-angle limit: pendulum SHM is only an approximation; for $\theta >15°$ the period is larger than $2\pi \sqrt{L/g}$. AP problems stay in small-angle regime.
• Amplitude and period: in SHM the period does not depend on amplitude. Unlike a pendulum at large angles, where amplitude does affect period.
• Energy in SHM: total energy is $\frac{1}{2}k{A}^2$, not \tfrac{1}{2}mv_\max^2. They're equal in magnitude but you should derive from amplitude when given.

7. Practice Questions (CED Style)

1. Two pendulums of lengths 1.0 m and 4.0 m oscillate on Earth. Find the ratio of their periods.
2. A 0.20 kg mass on a spring oscillates with period 0.50 s. Find the spring constant. If the amplitude is 0.05 m, find v_\max.
3. A pendulum is taken from Earth ($g=9.8$ m/s²) to the Moon ($g=1.6$ m/s²). By what factor does its period change?

8. Quick Reference Cheatsheet

• Restoring force: $F=-kx$ (defines SHM).
• Spring period: $T=2\pi \sqrt{m/k}$.
• Pendulum period: $T=2\pi \sqrt{L/g}$ (small angles).
• Position: $x(t)=A\cos (\omega t+\varphi )$.
• v_\max = A\omega at equilibrium; a_\max = A\omega^2 at extrema.
• Total energy = $\frac{1}{2}k{A}^2$.
• Period independent of amplitude (in true SHM).

9. What's Next

SHM connects forward to waves and sound (extensions in Phys 2) and fluids (waves in fluids). It's also the conceptual foundation for AC circuits in AP Phys 2 and quantum oscillators later. Use Ollie to walk through pendulum-on-cart, vertical-spring (with gravity offset), or coupled-oscillator problems.