Chapter 15, Problem 1OQ

To determine

Change in frequency of simple pendulum on increasing its length by two times.

Answer to Problem 1OQ

Option (d). It becomes $\frac{1}{\sqrt{2}}$ times as large.

Explanation of Solution

Write the equation to find the time period of simple pendulum.

    $T=2\pi \sqrt{\frac{l}{g}}$

Here, $T$ is the time period, $l$ is the length of pendulum, and $g$ is the gravitational acceleration.

Write the relation between frequency and $T$.

    $f=\frac{1}{T}$

Here,$f$ is the frequency.

Rewrite the above expression by substituting $T=2\pi \sqrt{\frac{l}{g}}$ for $T$.

    $f=\frac{1}{2\pi }\sqrt{\frac{g}{l}}$

Rewrite the above equation to find the new frequency.

    $f_{2l}=\frac{1}{2\pi }\sqrt{\frac{g}{l^{\prime }}}$

Here, $f_{2l}$ is the frequency of simple pendulum on increasing its length by two times and $l^{\prime }$ is the new length.

Rewrite the above equation by substituting $2l$ for $l^{\prime }$.

  $\begin{array}{c}{f}_{2l}=\frac{1}{2\pi }\sqrt{\frac{g}{2l}}\\ =\frac{1}{\sqrt{2}}\left(\frac{1}{2\pi }\sqrt{\frac{g}{\ell }}\right)\\ =\frac{f}{\sqrt{2}}\end{array}$

Conclusion:

It is found that as length of pendulum is increased by two times, frequency of oscillation gets decreases by $\frac{1}{\sqrt{2}}$ times. Therefore, option (d) is correct.

Therefore, options (a), (b), (c), and (e) are incorrect.