Question

A pendulam has a time period $T$ for small oscillations. An obstacle $\text{P}$ is situated below the point of suspension $\text{O}$ at a distance $\frac{3l}{4}$ as shown in the figure below. The pendulum is released from rest. Throughout the motion the moving string makes small angle with vertical. Find the time after which the pendulum returns back to its initial position.

• A. $T$
• B. $\frac{3T}{4}$
• C. $\frac{4T}{3}$
• D. $\frac{5T}{4}$

Answer 1

The correct option is B $\frac{3T}{4}$

A pendulum has time period $T$ for small oscillations.

Let, $T_1$ be the time period for length $l$. So,

$T_1=2\pi \sqrt{\frac{l}{g}}$

Now, the obstacle acts as point of suspension and $T_2$ be the time period for length $\frac{l}{4}$ with obstacle $\text{P}$, So

$T_2=2\pi \sqrt{\frac{l}{4g}}=\pi \sqrt{\frac{l}{g}}$


The total time period of oscillations or time after which the pendulum returns back is

$T=\frac{T_1}{2}+\frac{T_2}{2}$

$\Rightarrow T=\frac{2\pi \sqrt{\frac{l}{g}}}{2}+\frac{\pi \sqrt{\frac{l}{g}}}{2}$

$\Rightarrow T=\pi \sqrt{\frac{l}{g}}+\frac{\pi }{2}\sqrt{\frac{l}{g}}$

$\Rightarrow T=\frac{3\pi }{2}\sqrt{\frac{l}{g}}$

$\Rightarrow T=\frac{3}{4}\left[2\pi \sqrt{\frac{l}{g}}\right]$

Since, $T_1=2\pi \sqrt{\frac{l}{g}}$

$\therefore T=\frac{3T_1}{4}$

Hence, option (b) is the correct answer.