The pendulum.

Step 1 of 6

A longer string

A mass on a string, a pendulum, swings back and forth about its lowest point. Here the restoring force is gravity: when the bob is to one side, the part of gravity along the arc pulls it back towards the bottom. At small angles that part is almost in proportion to the distance from the bottom, so a pendulum too moves, to a good approximation, in simple harmonic motion.

Predict what will happen to the period when the string is made four times longer, from 25 cm to a metre, then watch the timer.

Step 2 of 6

The formula

When the string is tilted at an angle $\theta$⁠, the part of gravity along the arc is $mg\sin\theta$⁠. At small angles, in radians, $\sin\theta \approx \theta$$\sin\theta \approx$$\theta$⁠, and the distance from the bottom along the arc is $s = L\theta$⁠. So the restoring force is about

$F \approx -\frac{mg}{L}s$$F \approx$$-\frac{mg}{L}s$⁠

That is exactly Hooke’s law, with $\dfrac{mg}{L}$⁠ in place of $k$⁠. Put it into the spring’s formula:

$T = 2\pi\sqrt{\frac{m}{mg/L}} = 2\pi\sqrt{\frac{L}{g}}$$T =$$2\pi\sqrt{\frac{m}{mg/L}} =$$2\pi\sqrt{\frac{L}{g}}$⁠

The timer is now hidden. Work out the period of a pendulum 1.5 m long.

Step 3 of 6

A heavier bob

The mass cancelled in the formula. Predict whether that holds in the Scene too: what will happen to the period when the 0.5 kg bob is replaced by a 2 kg one?

Step 4 of 6

On the Moon

Gravity on the Moon is about 1.62 m/s², roughly a sixth of the Earth’s. It is now fixed at Moon, and the timer is hidden. Work out the period of the one-metre pendulum.

Step 5 of 6

A clock on the Moon

An astronaut wants to build a pendulum clock on the Moon whose every full swing takes 4.00 s. How long must the string be? Work it out, then move String length and check on the timer.

Step 6 of 6

Wide swings

The formula $T = 2\pi\sqrt{L/g}$⁠ rests on the approximation $\sin\theta \approx \theta$$\sin\theta \approx$$\theta$⁠. At $10^\circ$⁠ it is very good: $10^\circ$⁠ is about $0.1745$⁠ radians, and $\sin 10^\circ \approx 0.1736$$\sin 10^\circ \approx$$0.1736$⁠. At $80^\circ$⁠ it is not: $80^\circ$⁠ is about $1.396$⁠ radians, but $\sin 80^\circ \approx 0.985$$\sin 80^\circ \approx$$0.985$⁠. The Scene works out the full equation of motion, without the approximation.

Predict what will happen to the period when the bob is released from $80^\circ$⁠ instead of $10^\circ$⁠.