Pendulums oscillators and resonance

A pendulum is an oscillating system whose restoring force comes from gravity: when the bob is displaced from equilibrium, a force appears that points back toward equilibrium.

For small angles (roughly up to $10^\circ$) the period of a simple pendulum is $T=2\pi\sqrt{L/g}$ and is independent of the bob mass and nearly independent of amplitude — this property is called isochronism.

The total mechanical energy $E=\tfrac12 mv^2 + mgh$ is conserved when friction is neglected: at the turning points it is purely potential, and at the lowest point it is purely kinetic. The quantity $\omega_0=\sqrt{g/L}$ is the natural frequency of the system and sets its preferred rhythm. If an external periodic force drives the system at frequency $\omega$, then as $\omega \to \omega_0$ the amplitude of the forced oscillation grows sharply — this is resonance. Resonance creates no energy: it only means that the work done by the driving force each cycle is maximal, so the energy input rate matches the loss rate. Friction (damping) limits the height of the resonance peak and shifts it slightly toward lower frequencies.

A simple pendulum has length $L = 1.00$ m. Take $g = 9.81$ m/s². Compute the period $T$ and the natural frequency $f_0$. Will the period change if the bob mass is doubled?

A bob of mass $m = 0.40$ kg on a string of length $L = 1.50$ m is pulled aside to an angle $\theta = 10.0^\circ$ from the vertical and released from rest. Find the speed of the bob at the lowest point and the period of oscillation. Neglect air resistance.

Four pendulums hang from a common horizontal rod; their string lengths are $0.25$ m, $0.50$ m, $1.00$ m, and $2.00$ m. The $1.00$ m pendulum is set swinging, and the motion is weakly transmitted to the others through the rod. Which of the remaining pendulums will respond with the largest amplitude? Justify your answer with natural frequencies. How could you demonstrate true resonance in this system? Explain the role of damping.

Goal: measure the natural frequency of a pendulum and observe a resonance response without hazardous equipment.

Part 1 (home pendulum, safe). Tie a small mass (a coin, key, or nut up to $50$ g) to a light string about $1.0$ m long and fix the string with tape or a hook so that the bob hangs at least $10$ cm above the floor and away from your face and furniture. Pull the bob aside by a small angle (about $10^\circ$) and measure the time $N = 10$ complete oscillations. Compute $T = t/N$ and $f_0 = 1/T$, and compare with $2\pi\sqrt{L/g}$. Repeat with a different string length and plot $T^2$ against $L$ — the points should fall on a straight line with slope $4\pi^2/g$.

Part 2 (simulation for resonance). A real driven pendulum requires a driver and heavy loads, so study resonance in a simulation (for example PhET "Pendulum Lab" or "Resonance"). Set weak damping, sweep the driving frequency from $0.2$ to $1.5$ Hz in steps of $0.1$ Hz, wait for steady oscillations each time, and record the amplitude. Plot amplitude versus frequency, mark the peak, and compare its position with the calculated $f_0$. Then increase the damping and repeat: describe what happens to the height and width of the peak.

Safety: do not use heavy weights, do not swing the pendulum near your face, glass, or sharp objects; model large-amplitude forced oscillations on screen only.

Show the link between the mechanical scheme of an oscillator and its frequency response: where the natural frequency lies and how damping limits the resonance peak.

Diagram and graph. On the left, a pendulum of length $L$ with the bob displaced by an angle $\theta$; the turning points, the lowest point, and the height $h = L(1-\cos\theta)$ are marked, with arrows for gravity and the restoring component. On the right, a graph of the steady-state amplitude $A$ versus the driving frequency $f$: the natural frequency $f_0$ is marked on the $f$ axis; the curve with a sharp tall peak corresponds to weak damping, the lower and broader curve to strong damping; both peaks are shifted slightly to the left of $f_0$.

A teenager swings on a swing whose ropes are $L = 2.5\ \text{m}$ long. In one full cycle (forward and back) the seat travels about $1.0\ \text{m}$ with a maximum angle of roughly $12^\circ$ from the vertical. The period is

$$T = 2\pi\sqrt{\frac{L}{g}} = 2\pi\sqrt{\frac{2.5}{9.81}} \approx 3.2\ \text{s},$$

so the frequency is $f = 1/T \approx 0.31\ \text{Hz}$ — about three swings every ten seconds. Put a little brother of half the mass on the same swing and the period stays the same: both swing in step.

This is the small-oscillation regime of a pendulum: the restoring force is proportional to the displacement ($F \approx -\frac{mg}{L}x$), the motion is sinusoidal, and the period $T = 2\pi\sqrt{L/g}$ depends on neither the mass nor the amplitude.

A car body rests on four springs; one wheel carries a mass of $m = 350\ \text{kg}$ with a spring stiffness of $k = 25\,000\ \text{N/m}$. The natural frequency of the suspension is

$$f_0 = \frac{1}{2\pi}\sqrt{\frac{k}{m}} = \frac{1}{2\pi}\sqrt{\frac{25\,000}{350}} \approx 1.35\ \text{Hz}.$$

After a $0.10\ \text{m}$ speed bump the body bounces at this frequency, and the amplitude dies away in 2–3 oscillations, that is in $1.5$–$2.2\ \text{s}$.

This is the mass-on-a-spring harmonic oscillator: energy alternates between the elastic form $\frac{1}{2}kx^2$ and the kinetic form $\frac{1}{2}mv^2$, the frequency is set by the ratio $k/m$, and the weak friction in the shock absorbers provides the damping.

The drum with the laundry has a mass of $m = 20\ \text{kg}$ and hangs on springs of total stiffness $k = 8\,000\ \text{N/m}$, so its natural frequency is

$$f_0 = \frac{1}{2\pi}\sqrt{\frac{k}{m}} = \frac{1}{2\pi}\sqrt{400} \approx 3.2\ \text{Hz}$$

or $190\ \text{rpm}$. While accelerating to $1\,200\ \text{rpm}$ ($20\ \text{Hz}$) the machine passes through $190\ \text{rpm}$, and that is exactly when the cabinet shakes hardest and wanders across the bathroom floor.

This is resonance: the uneven laundry supplies a driving force at the rotation frequency of the drum, and when that frequency matches the natural frequency $f_0$ of the drum-and-springs system the amplitude of the forced oscillations rises sharply, then falls again once the drive moves away from resonance.