Question

The time period of a simple pendulum on the surface of the Earth is \( T \). At what height above the surface will the time period become \( 2T \)?
(Radius of Earth = \( 6400 \, \text{km} \))

• A. \( 3200 \, \text{km} \)
• B. \( 6400 \, \text{km} \)
• C. \( 1600 \, \text{km} \)
• D. \( 800 \, \text{km} \)

Hint:

For pendulum height problems, apply \( T_h = T \cdot \left(1 + \frac{h}{R}\right) \) when \( g \) changes with height.

Answer 1

The Correct Option is B

Time period \( T = 2\pi \sqrt{\frac{l}{g}} \Rightarrow T \propto \frac{1}{\sqrt{g}} \)
At height \( h \), \( g_h = \frac{g}{\left(1 + \frac{h}{R}\right)^2} \Rightarrow T_h = T \cdot \left(1 + \frac{h}{R}\right) \)
Given: \( T_h = 2T \Rightarrow 2 = 1 + \frac{h}{R} \Rightarrow \frac{h}{R} = 1 \Rightarrow h = R = 6400 \, \text{km} \)