Question

The time period of a simple pendulum on the surface of the earth is T. If the pendulum is taken to a height equal to half of the radius of the earth, then its time period is

• A. T/2
• B. 3T/2
• C. 2T
• D. 3T

Hint:

$T = 2\pi\sqrt{\frac{l}{g}}$. Gravity at height h: $g' = g(\frac{R}{R+h})^2$.

Answer 1

The Correct Option is C

The time period of a simple pendulum is given by $T = 2\pi\sqrt{\frac{l}{g}}$, where $l$ is the length of the pendulum and $g$ is the acceleration due to gravity. At the surface of the earth, $T = 2\pi\sqrt{\frac{l}{g}}$. At a height $h = R/2$ above the surface, the acceleration due to gravity is given by $g' = g(\frac{R}{R+h})^2 = g(\frac{R}{R+R/2})^2 = g(\frac{2}{3})^2 = \frac{4}{9}g$. The new time period $T' = 2\pi\sqrt{\frac{l}{g'}} = 2\pi\sqrt{\frac{l}{\frac{4}{9}g}} = 2\pi\sqrt{\frac{9l}{4g}} = \frac{3}{2}(2\pi\sqrt{\frac{l}{g}}) = \frac{3}{2}T$. $T' = 2\pi\sqrt{\frac{l}{g'}} = 2\pi \sqrt{\frac{l}{g\frac{R^2}{(R+R/2)^2}}} = 2\pi \sqrt{\frac{l}{g}\frac{9R^2}{4R^2}} = \frac{3}{2}2\pi\sqrt{\frac{l}{g}}=\frac{3}{2}T$ However, the given options are incorrect and does not contain the $\frac{3}{2}T$ term. At height $h$, $g_h = \frac{gR^2}{(R+h)^2}$. If $h=R/2$ then $g_h = g\frac{4}{9}$. Then $T' = 2\pi \sqrt{\frac{l}{\frac{4g}{9}}}=\frac{3}{2}(2\pi\sqrt{\frac{l}{g}}) = \frac{3}{2}T$ If the pendulum is taken to a height $h=R/2$, where $R$ is the Earth's radius. The acceleration due to gravity at a height $h$ is given by $g' = g(1+\frac{h}{R})^{-2}$. If we use $g'=g(\frac{R}{R+h})^2 = \frac{4g}{9}$. The new time period will be $T' = 2\pi\sqrt{\frac{l}{g'}} = 2\pi\sqrt{\frac{l}{\frac{4g}{9}}} = \frac{3}{2} 2\pi\sqrt{\frac{l}{g}}=\frac{3}{2}T$. If we consider $g' = \frac{GM}{(R+h)^2} = \frac{GM}{(1.5R)^2} = \frac{4}{9}\frac{GM}{R^2} = \frac{4g}{9}$ then $T'=\frac{3T}{2}$.