Question

When the length of a simple pendulum is increased by $36cm$, its time period of oscillation is found to increase by $25\mathrm{\%}$. The initial length of the simple pendulum (Acceleration due to gravity is $9.8m{s}^{-2}$):
$\begin{array}{rl}& \text{A}\text{. }36cm\\ & \text{B}\text{. }25cm\\ & \text{C}\text{. }64cm\\ & \text{D}\text{. }46cm\end{array}$

Answer 1

Hint: Time period of a simple pendulum is defined as the time taken by a pendulum to complete one full oscillation and Amplitude of pendulum is the distance travelled by pendulum from equilibrium position to one extreme side.

Complete step by step answer:
A pendulum is a weight, a massive bob, suspended from a pivot so that it can swing freely under the force of gravity. The length of an ideal simple pendulum is the distance from the pivot point to the center of mass of the bob. The radius of oscillation of a physical pendulum is related to the length of the pendulum, the mass of the pendulum, and the moment of inertia of the pendulum about the pivot point. Time period of a simple pendulum is defined as the time taken by a pendulum to complete one full oscillation
Time period of oscillation of a simple pendulum is given by the formula $T=2\mathrm{\Pi }\sqrt{{\displaystyle \frac{l}{g}}}$
Where,
$l$ is the length of string of pendulum
$g$ is the acceleration due to gravity
Now,
We are given that, when the length of a simple pendulum is increased by $36cm$, its time period of oscillation is found to increase by $25\mathrm{\%}$. We have to calculate the initial length of the simple pendulum.
Let the initial time period of simple pendulum,
$T=2\pi \sqrt{{\displaystyle \frac{l}{g}}}$
Now,
New length of the pendulum,
$l^{\prime }=l+{\displaystyle \frac{36}{100}}$
New time period of the pendulum,
$T^{\prime }=T+{\displaystyle \frac{25}{100}}T=1.25T$
Therefore,
$T^{\prime }=2\pi \sqrt{{\displaystyle \frac{l^{\prime }}{g}}}$
Or,
$1.25T=2\pi \sqrt{{\displaystyle \frac{l+0.36}{g}}}$
Put, $T=2\pi \sqrt{{\displaystyle \frac{l}{g}}}$
$\begin{array}{rl}& \Rightarrow 1.25\times 2\pi \sqrt{{\displaystyle \frac{l}{g}}}=2\pi \sqrt{{\displaystyle \frac{l+0.36}{g}}}\\ & \Rightarrow 1.25\sqrt{{\displaystyle \frac{l}{g}}}=\sqrt{{\displaystyle \frac{l+0.36}{g}}}\end{array}$
Squaring both sides,
$\begin{array}{rl}& \Rightarrow 1.5625\times {\displaystyle \frac{l}{g}}={\displaystyle \frac{l+0.36}{g}}\\ & \Rightarrow 1.5625l=l+0.36\\ & \Rightarrow l(1.5625-1)=0.36\\ & \Rightarrow 0.5625l=0.36\\ & \Rightarrow l={\displaystyle \frac{0.36}{0.5625}}=0.64\\ & \Rightarrow l=0.64m=64cm\end{array}$
The initial length of the simple pendulum is $64cm$

So, the correct answer is “Option C”.

Note: Always remember that the time period of oscillation of a simple pendulum does not depend upon the mass of the oscillating object and the amplitude of the oscillation. It only depends upon the length of the pendulum and the acceleration due to gravity.