pendulum periods and oscillation energy

A simple pendulum is a bob of mass $m$ on a massless string of length $L$; at an angle $\theta$ the restoring force is $-mg\sin\theta$.

For small angles $\sin\theta\approx\theta$, so $F\approx-\frac{mg}{L}x$ — a quasi-elastic force with stiffness $k=mg/L$, and from the harmonic-oscillator dynamics $T=2\pi\sqrt{m/k}$ it follows that $T=2\pi\sqrt{\frac{L}{g}}.$

The period contains neither the mass nor (at small angles) the amplitude: it is set only by the string length and the free-fall acceleration. The oscillation energy is described by the same model: $E=\tfrac12 kA^2=\tfrac12 m\omega^2 A^2$, where the amplitude along the arc is $A=L\theta_0$ and $\omega=2\pi/T$. During the motion energy passes from potential to kinetic and back, while the total energy stays constant: $E=mgL(1-\cos\theta_0)\approx\tfrac12 mgL\theta_0^2$. The maximum speed occurs at the lowest point: $v_{\max}=\omega A=\theta_0\sqrt{gL}$; above roughly $15^\circ$ the approximation $\sin\theta\approx\theta$ worsens and the period starts to grow with amplitude.

A simple pendulum has length $L=1.00$ m. Find the period of its small oscillations for $g=9.81$ m/s². Does the answer depend on the bob mass?

A pendulum of length $L=0.50$ m with a bob of mass $m=0.20$ kg is released from $\theta_0=8.0^\circ$ ($0.140$ rad). Find the arc amplitude, the maximum speed, and the total oscillation energy.

A pendulum of length $L=1.00$ m is released from $\theta_0=6.0^\circ$.

(a) Using energy conservation, find the speed of the bob when the string makes an angle $\theta=3.0^\circ$ with the vertical.

(b) By what factor does the period change if the same pendulum is taken to the Moon, where $g_{\text{M}}=1.62$ m/s²?

Simulation (safe self-study plan): open the PhET Pendulum Lab or an equivalent simulation.

1. Set $L=1.00$ m, $m=0.20$ kg, $\theta_0=5^\circ$. Time 10 full oscillations and find $T$.
2. Repeat for $L=0.25,\ 0.50,\ 1.00,\ 2.00$ m. For each value compute $\sqrt{L}$ and plot $T$ versus $\sqrt{L}$ — you should get a straight line through the origin.
3. At fixed length vary the mass ($0.05,\ 0.20,\ 1.00$ kg) and confirm that the period does not change.
4. Vary $\theta_0=5^\circ,\ 15^\circ,\ 45^\circ$ and note at which angle the measured period noticeably exceeds $2\pi\sqrt{L/g}$.
5. Turn on the energy display and describe how kinetic and potential energy exchange over one period: where $E_k$ is maximal, where $E_p$ is maximal, and what their sum equals.

Show that the period of small oscillations is set only by the length (the graph of $T$ versus $\sqrt{L}$ is a straight line) and illustrate the exchange of kinetic and potential energy within one period.

Sketch and graphs.

1. On the left, the pendulum in three positions: far left ($\theta_0$, $v=0$), bottom ($0$, $v=v_{\max}$), far right ($-\theta_0$, $v=0$); arrows show the velocity and the restoring force $mg\sin\theta$.
2. In the middle, energy bars for these three points: at the extremes all energy is potential ($E_p=E$), at the bottom all is kinetic ($E_k=E$); the sum of the bars stays constant.
3. On the right, a graph of the period $T$ versus $\sqrt{L}$: the data points fall on the straight line $T=\frac{2\pi}{\sqrt{g}}\sqrt{L}$ with slope $2\pi/\sqrt{g}$.

A child of mass $30\ \text{kg}$ sits on a swing with suspension length $L=2.5\ \text{m}$. The child is pulled back so the arc length from the lowest point is $0.60\ \text{m}$ and released. The deflection angle $\theta = s/L = 0.60/2.5 = 0.24\ \text{rad}$ is small, so the period is $T = 2\pi\sqrt{L/g} = 2\pi\sqrt{2.5/9.81} = 3.17\ \text{s}$. Ten full oscillations take $31.7\ \text{s}$. The rise height is $h = L(1-\cos\theta) = 2.5(1-\cos 0.24) = 0.0717\ \text{m}$, the maximum speed is $v_{\max} = \sqrt{2gh} = 1.19\ \text{m/s}$, and the maximum kinetic energy is $E_k = \frac12 m v_{\max}^2 = 21.1\ \text{J}$; it equals the maximum potential energy $E_p = mgh = 21.1\ \text{J}$.

This exercises the simple-pendulum period formula $T=2\pi\sqrt{L/g}$ and conservation of mechanical energy $E_k + E_p = \text{const}$; the period is independent of mass and, for small angles, of amplitude.

A grandfather clock pendulum is adjusted to a period $T=2.00\ \text{s}$, so each half-oscillation lasts $1.00\ \text{s}$. From $T=2\pi\sqrt{L/g}$ the length is $L = g(T/2\pi)^2 = 9.81(2/(2\pi))^2 = 0.994\ \text{m}$. The bob mass is $m=0.50\ \text{kg}$, and the amplitude is $\theta_0 = 2.0^\circ = 0.0349\ \text{rad}$. The rise height is $h = L(1-\cos\theta_0) = 0.994(1-\cos 2^\circ) = 6.06\cdot10^{-4}\ \text{m}$, the maximum potential energy is $E_p = mgh = 2.97\cdot10^{-3}\ \text{J} = 2.97\ \text{mJ}$, the maximum speed is $v_{\max} = \sqrt{2gh} = 0.109\ \text{m/s}$, and the maximum kinetic energy is equal: $E_k = \frac12 m v_{\max}^2 = 2.97\ \text{mJ}$. Friction and air-drag losses each half-oscillation are replaced by the falling drive weight.

This applies $T=2\pi\sqrt{L/g}$ and $h = L(1-\cos\theta_0)$ together with the $E_p \leftrightarrow E_k$ conversion in a pendulum; the small angle $2^\circ$ justifies the harmonic approximation.

A wrecking ball of mass $m=1200\ \text{kg}$ hangs on a cable of length $L=15\ \text{m}$ and is pulled aside by an angle $\theta_0 = 10.0^\circ$. Its period is $T = 2\pi\sqrt{L/g} = 2\pi\sqrt{15/9.81} = 7.77\ \text{s}$. The rise height is $h = L(1-\cos 10^\circ) = 15(1-0.9848) = 0.228\ \text{m}$, and the maximum potential energy is $E_p = mgh = 1200\cdot9.81\cdot0.228 = 2.68\cdot10^3\ \text{J} = 2.68\ \text{kJ}$. At the lowest point the speed is $v_{\max} = \sqrt{2gh} = 2.12\ \text{m/s}$, and the kinetic energy is $E_k = \frac12 m v_{\max}^2 = 2.68\ \text{kJ}$; this energy is transferred to the wall on impact.

The same period formula $T=2\pi\sqrt{L/g}$ and energy conservation are used; the period depends only on $L$ and $g$, not on $m$, while the impact energy is set by the height $h=L(1-\cos\theta_0)$.