Question

A small bob attached to a light inextensible thread of length $l$ has a periodic time $T$ when allowed to vibrate as a simple pendulum. The thread is now suspended from a fixed end O of a vertical rigid rod of length $\frac{3l}{4}$ (as in figure.) If now the pendulum performs periodic oscillations in this arrangement, the periodic time will be

• A. $\frac{3T}{4}$
• B. $\frac{T}{2}$
• C. T
• D. 2T

Answer 1

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

We have
$T=2\pi \sqrt{\frac{l}{g}}$
Now the rod is introduced in the pendulum. This will make pendulum roatate about one end of the rod for half of the cycle with length of $\frac{l}{4}$ and in the remaining half cycle pendulum will rotate about the complete $l$ length. So, we can divide the time period $T_{new}$ in $T_{1}and{T}_2$
$T_1$= pendulum with $\frac{l}{4}$ length i.e., first half cycle.
$T_2$= pendulum with $l$ length i.e., second half cycle.
Clearly, $T_2=\frac{T}{2}$
and $T_1=\frac{1}{2}\left(2\pi \sqrt{\frac{\frac{l}{4}}{g}}\right)=\frac{1}{2}\left(2\pi \sqrt{\frac{l}{4g}}\right)$
$=\frac{1}{4}\left(2\pi \sqrt{\frac{l}{g}}\right)=\frac{T}{4}$
$\Rightarrow T_{new}=T_1+T_2=\frac{T}{4}+\frac{T}{2}=\frac{3T}{4}$