The simple pendulum

The restoring torque

Narrated video. Transcript:

Tie a small heavy stone to a thread and hang it from a fixed support. This is a simple pendulum: a point mass m, called the bob, on a light, inextensible string of length L.

Pull the bob aside so the string makes an angle θ with the vertical, and let go. It swings to and fro in a vertical plane about its lowest point. Is this simple harmonic motion?

Look at the forces on the bob. There are just two: the tension T along the string, and the weight m g straight down.

Split the weight into two parts. m g cos θ acts along the string, and m g sin θ acts along the arc, at right angles to the string.

Along the string, the tension and m g cos θ together give the centripetal force for the circular path. Both act along the string, through the support, so they produce no torque about it.

Only m g sin θ has a torque about the support. Its lever arm is L, so the torque is τ equals minus L m g sin θ. The minus sign shows that it always turns the bob back towards the vertical: a restoring torque.

Now use the rotational form of Newton’s second law, τ equals I α. For a point bob, the moment of inertia about the support is I equals m L squared.

So m L squared α equals minus m g L sin θ. Cancel m L: the angular acceleration is α equals minus g over L times sin θ.

Compare with SHM, where the acceleration is minus ω squared times the displacement. Here we have sin θ, not θ. So, strictly, a pendulum is not simple harmonic. Unless the angle is small.

Small angles and the period

Narrated video. Transcript:

Plot sin θ and θ in radians on the same graph. Near zero, the two curves lie almost on top of each other.

At 5 degrees, θ is 0.0873 radians and sin θ is 0.0872. Even at 20 degrees, 0.349 against 0.342: only 2 percent apart. For small angles, sin θ is approximately θ.

With this small-angle approximation, α equals minus g over L times θ. That is exactly the SHM form, with ω squared equal to g over L.

So the angular frequency is ω equals the square root of g over L, and the period of a simple pendulum is T equals 2π times the square root of L over g.

The mass of the bob has cancelled out. A brass bob and a wooden bob on the same length of string swing with the same period. For small swings, the amplitude does not matter either.

Legend has it that young Galileo noticed this in a cathedral, timing a swinging lamp against his own pulse. Big swings and small swings seemed to take the same time.

Worked example: what length of pendulum ticks seconds, with a period of 2 seconds? L equals g T squared over 4π squared: 9.8 times 4, over 39.5, which is about 0.99 metres. Almost exactly one metre.

Take the same pendulum to the Moon, where g is about one sixth as large. T depends on one over root g, so it swings root 6, about 2.45 times, more slowly.

For large swings, sin θ is noticeably smaller than θ, so the restoring torque is weaker and the real period is longer than 2π root L over g. At 60 degrees it is already about 7 percent longer.

Try it: pendulum lab

Interactive simulation. Things to try:

• Press Release (Space). The dashed blue twin follows the small-angle formula exactly.
• At 10° the two stay together for many swings. Raise θ₀ to 60° and watch the real one fall behind.
• Change the mass: nothing about the motion changes.
• Try the Moon and Jupiter. Which way does T change, and by how much?

A lighter bob

The brass bob of a simple pendulum is replaced by a wooden bob of the same size but one-eighth the mass. For small swings, the period:

1. becomes √8 times longer
2. becomes √8 times shorter
3. stays the same
4. becomes 8 times shorter

A one-second pendulum

What length of simple pendulum has a period of 1 s on the Earth (g = 9.8 m/s²)?

1. 0.25 m
2. 0.50 m
3. 1.0 m
4. 1.56 m

Summary

• Restoring torque: τ = −mgL sin θ
• Equation of motion: α = −(g/L) sin θ ≈ −(g/L) θ
• Period (small angles): T = 2π√(L/g)
• Seconds pendulum (T = 2 s): L ≈ 1 m

• Only the component mg sin θ of the weight has a torque about the support. It always pulls the bob back towards the vertical.
• For small angles, sin θ ≈ θ (in radians), so the pendulum performs SHM with ω = √(g/L).
• The period depends only on L and g, not on the mass of the bob, and not on the amplitude for small swings.
• For large amplitudes, the real period is longer than 2π√(L/g): about 7% longer at 60°.