Simple harmonic motion and the pendulum

Physics · Senior school physics · about 10 min

A swinging pendulum provides a familiar model of oscillation. Near its equilibrium position, a small displacement produces a restoring acceleration directed back towards equilibrium. This leads to simple harmonic motion, a model that connects force, frequency, period and energy.

The key idea

For an ideal simple pendulum swinging through small angles,

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

Here $T$ is period, $L$ is the distance from the pivot to the bob's centre of mass and $g$ is gravitational acceleration. The angle must be small enough for the approximation to apply.

Try it yourself

Change the mass, stiffness and damping, then watch the spring oscillate and die away.

Recognising simple harmonic motion

Simple harmonic motion occurs when the acceleration is proportional to displacement from equilibrium and directed towards it:

a=-\omega^2x.

The negative sign represents the restoring direction. A sinusoidal displacement model is $x=A\cos(\omega t+\phi)$ , where $A$ is amplitude, $\omega$ is angular frequency and $\phi$ is phase. The period and frequency satisfy $T=2\pi/\omega$ and $f=1/T$ . At the turning points the speed is zero and the restoring acceleration has its greatest magnitude. At equilibrium the speed is greatest and the restoring acceleration is zero.

The small-angle pendulum

For a bob on a light string of length $L$ , gravity has a tangential component $-mg\sin\theta$ . For a small angular displacement measured in radians, $\sin\theta\approx\theta$ . The rotational equation then becomes $\ddot\theta+(g/L)\theta=0$ , which has the form of simple harmonic motion with $\omega=\sqrt{g/L}$ . That is why the period depends on length and gravity, but not on the bob's mass. For large swings, the small-angle formula underestimates the true period. Air resistance and pivot friction gradually reduce the amplitude of a real pendulum.

One fixed pivot at the top with a vertical dashed equilibrium line descending from it to the lowest bob position. A string of length L extends from the pivot down and to the right at a small angle θ from the dashed vertical, ending at a round bob. Draw a downward gravitational force arrow from the bob labelled mg and a string-directed force arrow from the bob towards the pivot labelled F_T. Draw a curved arrow tangential to the arc pointing from the displaced bob towards the lower equilibrium point, labelled restoring direction. Label the arc of swing faintly and mark the lowest point equilibrium. Keep the displaced angle visibly small. — The tangential component of gravity restores a displaced pendulum towards equilibrium.

Period, frequency and predictions

The period is the time for one complete back-and-forth cycle, not the time to travel from one extreme to the other. Frequency counts cycles per second and has units $\mathrm{Hz}$ . To find the approximate number of cycles in a duration $t$ , use $n=t/T$ . Because $T\propto\sqrt L$ at fixed gravity, making a pendulum four times longer doubles its period. A heavier bob does not change the ideal small-angle period. Always convert length to metres and use consistent SI units before evaluating the square root.

Worked example: Timing a small-angle pendulum

A simple pendulum has length $0.81\,\mathrm{m}$ and swings through a small angle. Take $g=9.81\,\mathrm{m/s^2}$ . Calculate its period, its frequency and approximately how many complete oscillations it performs in $60\,\mathrm{s}$ . Neglect damping.

A pendulum hangs from a fixed horizontal support, with one pivot and a thin string sloping slightly right from the dashed vertical line. Label the string distance from pivot to the centre of the round bob 0.81 m. Mark a small angular displacement at the pivot and show a downward arrow outside the swing labelled g = 9.81 m/s². Draw a thin curved path arc through the equilibrium point with a short arrow from the rightmost displaced bob towards equilibrium. Do not show a calculated period, frequency, cycle count or answer. — The small-angle period depends on string length and local gravity, not bob mass.

Show the solution

Step 1: Calculate the period.

T=2\pi\sqrt{\frac{L}{g}}=2\pi\sqrt{\frac{0.81}{9.81}}=1.80546\ldots\,\mathrm{s}\approx1.81\,\mathrm{s}.

Step 2: Convert period to frequency. Use the unrounded period:

f=\frac1T=0.55388\ldots\,\mathrm{Hz}\approx0.554\,\mathrm{Hz}.

Step 3: Estimate the number of cycles.

n=\frac{60}{T}=33.233\ldots\text{ cycles}.

Because only complete oscillations count, $33$ are completed within the stated time.

Answer: period $1.81\,\mathrm{s}$ , frequency $0.554\,\mathrm{Hz}$ and $33$ complete oscillations in $60\,\mathrm{s}$ .

Common mistakes

Treating a half-swing as a complete period

Measure the time to return to the same position while moving in the same direction.

Using pendulum mass in the ideal period equation

For small-angle motion, period depends on $L$ and $g$ , not on bob mass.

Using the approximation for very large swings

Use the small-angle formula only when $\sin\theta\approx\theta$ in radians is justified.

Confusing angular frequency and ordinary frequency

Remember $\omega=2\pi f$ ; their units are $\mathrm{rad/s}$ and $\mathrm{Hz}$ .

Check your understanding

1. At the same gravitational acceleration, what happens to the small-angle pendulum period when its length doubles?

2. An oscillator has a frequency of $2.5\,\mathrm{Hz}$ . What is its period?

3. A pendulum bob of mass $0.20\,\mathrm{kg}$ is replaced by one of mass $0.40\,\mathrm{kg}$ . Length and small amplitude stay the same. What happens to the period?

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