Pendulums and Oscillation Energy

Driving question

Why does pendulum length set the clock while bob mass does not?

Before: prediction

Two small-angle pendulums have equal length but different bob masses. Predict which has the longer period, then identify what measurement could disprove your prediction.

Hook: a heavier bob does not make a slower clock

For an ideal simple pendulum of length $L$ at small angular amplitude,

$T=2\pi\sqrt{\frac{L}{g}}.$

Mass is absent because both the restoring gravitational effect and inertia scale with mass. Length matters: quadrupling $L$ doubles $T$.

Where the model comes from

Tangential gravity is $-mg\sin\theta$. With arc displacement $s=L\theta$,

$mL\frac{d^2\theta}{dt^2}=-mg\sin\theta.$

For small angles measured in radians, $\sin\theta\approx\theta$, so

$\frac{d^2\theta}{dt^2}=-\frac{g}{L}\theta.$

This has simple-harmonic form with $\omega=\sqrt{g/L}$ and $T=2\pi/\omega$.

Worked example: length from period

A small-angle pendulum has $T=1.60\text{ s}$. With $g=9.8\text{ m/s}^2$,

$L=g\left(\frac{T}{2\pi}\right)^2=9.8\left(\frac{1.60}{2\pi}\right)^2=0.636\text{ m}.$

Energy exchange

Choosing zero gravitational potential at the bottom, the exact potential energy is

$U=mgL(1-\cos\theta).$

At a turning point, speed is zero and energy is all potential. At the bottom, potential is minimum and speed is maximum. With negligible losses,

$E=K+U=\text{constant}.$

Worked example: bottom speed

Released from rest at $L=0.80\text{ m}$ and $\theta_0=20^\circ$,

$\frac12mv^2=mgL(1-\cos\theta_0),$

$v=\sqrt{2gL(1-\cos20^\circ)}\approx0.973\text{ m/s}.$

Mass cancels again.

Graph relationships

For ideal SHM, displacement is sinusoidal; velocity is shifted by one-quarter cycle; acceleration points opposite displacement. Kinetic energy and potential energy repeat twice per displacement cycle because they depend on squared quantities.

During: retrieval pause

At equilibrium, rank speed, acceleration magnitude, kinetic energy, and potential energy as maximum, minimum, or zero.

Damping and driving

Air resistance and pivot friction transfer mechanical energy to thermal energy, reducing amplitude. Weak damping often changes period only slightly while the envelope shrinks. Periodic driving can add energy; resonance occurs when driving frequency is near the system's natural frequency, though damping limits amplitude.

Misconception checks

• A pendulum's acceleration is not zero at the turning points; its speed is zero there.
• The small-angle period formula is an approximation, not exact at large amplitude.
• Doubling length multiplies period by $\sqrt2$, not 2.
• Mass cancels only under the ideal simple-pendulum assumptions.
• Energy frequency is twice displacement frequency.

After: transfer

A pendulum's measured period grows as the release angle increases from $5^\circ$ to $45^\circ$. Explain why this does not contradict the small-angle formula and propose a graph that reveals the model's range.

AP-style synthesis

State the system and zero-energy reference, identify whether the small-angle approximation applies, choose force/torque or energy reasoning, preserve radians in the approximation, and compare predictions with uncertainty.

Further learning

See sources.yaml. All prose, examples, questions, and diagrams are original and do not imply College Board endorsement.