Question

A simple pendulum doing small oscillations at a place R height above earth surface has time period of T1 = 4 s. T2 would be it's time period if it is brought to a point which is at a height 2R from earth surface. Choose the correct relation [R = radius of Earth]:

• A. T1 = T2
• B. 2T1 = 3T2
• C. 3T1 = 2T2
• D. 2T1 = T2

Answer 1

The Correct Option is C

The time period of a simple pendulum located at a height h = R above the Earth's surface is determined by the formula:

\[ T_1 = 2\pi \sqrt{\frac{\ell}{g_{\text{eff}}}} \]

Here, the effective acceleration due to gravity at height \(h = R\) is calculated as: \[ g_{\text{eff}} = \frac{GM}{(2R)^2} = \frac{g}{4} \]

Consequently, the time period at height \(R\) is: \[ T_1 = 2\pi \sqrt{\frac{\ell}{g/4}} = 2\pi \sqrt{\frac{4\ell}{g}} \]

Similarly, at a height \(h = 2R\), the effective acceleration due to gravity is: \[ g_{\text{eff}} = \frac{GM}{(3R)^2} = \frac{g}{9} \]

Thus, the time period \(T_2\) is: \[ T_2 = 2\pi \sqrt{\frac{\ell}{g/9}} = 2\pi \sqrt{\frac{9\ell}{g}} \]

The ratio of these time periods is: \[ \frac{T_1}{T_2} = \sqrt{\frac{4}{9}} = \frac{2}{3} \]

This relationship can be expressed as: \[ 3T_1 = 2T_2 \]