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 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: \[ 3T_1 = 2T_2 \]

Answer 2

Step 1: Write the time period formula at a height above the Earth’s surface
The time period of a simple pendulum depends on the effective acceleration due to gravity at that height: \[ T = 2\pi\sqrt{\frac{l}{g_h}}, \] where \( g_h = g\left(\frac{R}{R + h}\right)^2. \)

Step 2: Compare the two cases
At height \( h_1 = R \): \[ T_1 = 2\pi\sqrt{\frac{l}{g\left(\frac{R}{R + R}\right)^2}} = 2\pi\sqrt{\frac{l}{g\left(\frac{R}{2R}\right)^2}} = 2\pi\sqrt{\frac{l}{g/4}} = 4\pi\sqrt{\frac{l}{g}}. \] At height \( h_2 = 2R \): \[ T_2 = 2\pi\sqrt{\frac{l}{g\left(\frac{R}{R + 2R}\right)^2}} = 2\pi\sqrt{\frac{l}{g\left(\frac{R}{3R}\right)^2}} = 2\pi\sqrt{\frac{l}{g/9}} = 6\pi\sqrt{\frac{l}{g}}. \]

Step 3: Find the relation between \( T_1 \) and \( T_2 \)
From above: \[ T_1 = 4\pi\sqrt{\frac{l}{g}}, \quad T_2 = 6\pi\sqrt{\frac{l}{g}}. \] Thus, \[ 3T_1 = 2T_2. \]

Final answer

3T₁ = 2T₂