Question

When the length of a pendulum is doubled, its frequency will be cut in half. Is this true or false?

Answer 1

The period T of a pendulum with length L is given by the formula:

`T=2\pi\cdot\sqrt[]{(L)/(g)}`

Where g is the acceleration of gravity:

`g=9.81(m)/(s^2)`

If the length of the pendulum is doubled, the period will be:

`\begin{gathered} T^(\prime)=2\pi\cdot\sqrt[]{(2L)/(g)} \\ =\sqrt[]{2}*2\pi\cdot\sqrt[]{(L)/(g)} \\ =\sqrt[]{2}* T \end{gathered}`

On the other hand, the frequency is the reciprocal of the period. Then:

`\begin{gathered} f=(1)/(T) \\ f^(\prime)=(1)/(T^(\prime))=\frac{1}{\sqrt[]{2}* T}=\frac{1}{\sqrt[]{2}}*(1)/(T)=\frac{1}{\sqrt[]{2}}* f \end{gathered}`

Then, if the length of the pendulum is doubled, the frequency is cut by a factor of 1/√2:

`f^(\prime)=\frac{1}{\sqrt[]{2}}* f`

This is not the same as if the frequency was cut by a factor of 1/2.

Therefore, the statement is:

`\text{False}`