Question

If a simple pendulum oscillates with small amplitude and its length is doubled, what happens to the frequency of its motion?

Answer 1

The period of the simple pendulum can be written as follows:

$$T=2\pi\sqrt{\dfrac{L}{g}}.$$

If the length of the pendulum is doubled, the new period can be found as follows:

$$T_{new}=2\pi\sqrt{\dfrac{2L}{g}}=2\sqrt{2}\sqrt{\dfrac{2L}{g}}=\sqrt{2}T.$$

Finally, we can find the new frequency of the pendulum:

$$f_{new}=\dfrac{1}{T_{new}}=\dfrac{1}{\sqrt{2}T}.$$

Therefore, if a simple pendulum oscillates with small amplitude and its length is doubled the frequency of its motion becomes $\dfrac{1}{\sqrt{2}}$ times as large.