Question

A simple pendulum is taken to a place where the acceleration due to gravity falls to one-fourth. The time period of the pendulum becomes ___ its initial value.

• A. twice
• B. four times
• C. eight times
• D. half

Answer 1

Answer: (A)

twice

We have time period of a pendulum $T\propto \frac{1}{\sqrt{g}}$.

$\therefore \frac{T_1}{T_2}=\sqrt{\frac{g_2}{g_1}}$ where $T_1$ and $T_2$ are the initial and final time periods of the simple pendulum and $g_1$ and $g_2$ are the initial and final time acceleration due to gravity. If the acceleration due to gravity falls to one-fourth i.e.,$g_2=\frac{g_1}{4}$

$\Rightarrow \frac{T_1}{T_2}=\sqrt{\frac{g_1}{4g_1}}$ $\Rightarrow \frac{T_1}{T_2}=\frac{1}{2}$ $\Rightarrow T_2=2T_1$

i.e., the time period of the pendulum becomes twice its initial value.