Question #160903
If a simple pendulum oscillates with small amplitude and its length is doubled, what happens to the frequency of its motion?
1
Expert's answer
2021-02-15T00:59:03-0500

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


T=2πLg.T=2\pi\sqrt{\dfrac{L}{g}}.

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


Tnew=2π2Lg=222Lg=2T.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:


fnew=1Tnew=12T.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 12\dfrac{1}{\sqrt{2}} times as large.


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