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 "3l/4". If now the pendulum performs periodic oscillations in this arrangement, the periodic time will be:

• A. $3T/4$
• B. $4T/5$
• C. $2T/3$
• D. $5T/6$

Answer 1

The correct option is A $3T/4$

Half of the oscillation is completed with length $l$ and rest half with $\frac{l}{4}.$
$\therefore$ Time period$=\frac{T_1}{2}+\frac{T_2}{2}$
$=\frac{1}{2}[2\pi \sqrt{\frac{l}{g}}+2\pi \sqrt{\frac{l/4}{g}}]$
$=\frac{3}{4}\left[2\pi \sqrt{\frac{l}{g}}\right]=\frac{3}{4}T$