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

Answer: (C)

3T1 = 2T2

Answer 2

The time period of a simple pendulum at a height h = R above the earth’s surface is given by:

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

where the effective acceleration due to gravity at a height $h = R$ is: $g_{\text{eff}} = \frac{GM}{(2R)^2} = \frac{g}{4}$

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

Hence, 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 the time periods is: $\frac{T_1}{T_2} = \sqrt{\frac{4}{9}} = \frac{2}{3}$

This implies: $3T_1 = 2T_2$