Question

A simple pendulum doing small oscillations at a place \( R \) height above the Earth's surface has a time period of \( T_1 = 4 \, {s} \). \( T_2 \) would be its time period if it is brought to a point which is at a height \( 2R \) from the Earth's surface. Choose the correct relation [\( R \) = radius of Earth]:

• A. \( T_1 = T_2 \)
• B. \( 2T_1 = 3T_2 \)
• C. \( 3T_1 = 2T_2 \)
• D. \( 2T_1 = T_2 \)

Hint:

The time period of a simple pendulum depends on the gravitational acceleration, which varies with height above the Earth's surface.

Answer 1

The Correct Option is C

Step 1: {Time period of a simple pendulum}
\[ T = 2\pi \sqrt{\frac{L}{g}} \quad {and} \quad g = \frac{GM}{(R + h)^2} \] Step 2: {Relating time periods at different heights}
\[ \frac{T_1}{T_2} = \frac{R + h_1}{R + h_2} = \frac{R + R}{R + 2R} = \frac{2}{3} \Rightarrow 3T_1 = 2T_2 \] Thus, the correct relation is \( 3T_1 = 2T_2 \).