Question

A simple pendulum has a time period $T_1$ when on the earth's surface and $T_2$ when taken to a height $R$ above the earth's surface, where $R$ is the radius of the earth. The value of $\frac{T_2}{T_1}$ is:

• A. $4$
• B. $2$
• C. $1$
• D. $\sqrt{2}$

Answer 1

Time period of a simple pendulum $=2\pi \sqrt{\frac{l}{g}}$

$\Rightarrow g$ on earth's surface $=\frac{GM}{R^2}$

$g$ on a height $R$ above earth's surface $=\frac{GM}{{\left(2R\right)}^2}$

$\Rightarrow T_1$ (time period on earth's surface ) $=2\pi \sqrt{\frac{l\left(R^2\right)}{GM}}$

$T_2$ (time period at a height ${}^{\prime }{h}^{\prime }$ above earth's surface ) $=2\pi \sqrt{\frac{l{\left(2R\right)}^2}{GM}}$

$\Rightarrow \frac{T_2}{T_1}=\sqrt{4}=2.$

Hence, the answer is $2.$