Question

If a seconds pendulum on the earth is taken to a planet whose gravity is half of the gravity on earth, its time period on that planet is

• A. \(2\text{ sec}\)
• B. \(4\text{ sec}\)
• C. \(4\sqrt{2}\text{ sec}\)
• D. \(2\sqrt{2}\text{ sec}\)

Hint:

For a simple pendulum, \(T\propto \frac{1}{\sqrt{g}}\). If gravity becomes half, time period becomes \(\sqrt{2}\) times.

Answer 1

The Correct Option is D

A seconds pendulum has time period: \[ T=2\text{ sec}. \] The time period of a simple pendulum is: \[ T=2\pi\sqrt{\frac{l}{g}}. \] For the same pendulum, length \(l\) remains constant. Therefore: \[ T\propto \frac{1}{\sqrt{g}}. \] Let the gravity on earth be: \[ g. \] On the new planet, gravity is half: \[ g'=\frac{g}{2}. \] Now: \[ \frac{T'}{T}=\sqrt{\frac{g}{g'}}. \] Substitute \(g'=\frac{g}{2}\): \[ \frac{T'}{T}=\sqrt{\frac{g}{g/2}}. \] \[ \frac{T'}{T}=\sqrt{2}. \] So: \[ T'=T\sqrt{2}. \] Since: \[ T=2\text{ sec}, \] we get: \[ T'=2\sqrt{2}\text{ sec}. \] Hence, the time period on that planet is: \[ 2\sqrt{2}\text{ sec}. \]