Question

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

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

Answer 1

The correct option is B

$\frac{3T}{4}$

After, P, length of pendulum becomes $\frac{1}{4}$

Now, as $T\propto \sqrt{l},$ so after P, time period will become $T^{\prime }=\frac{T}{2}.$ Therefore, the desired time will be :
$t=\frac{T}{2}+\frac{T^{\prime }}{2}=\frac{T}{2}+\frac{T}{4}=\frac{3T}{4}$