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

Answer 1

The correct option is D $2$

The time period of a simple pendulum depends upon the acceleration due to gravity

$T=2\pi \sqrt{\frac{L}{g}}$

Assuming $L$ is constant, then

$T\propto \frac{1}{\sqrt{g}}.....\left(1\right)$

Acceleration due to gravity at a height $h$ above the surface of the earth is given as

$g_h=\frac{g}{{(1+\frac{h}{R})}^2}$

at $h=R$

$g_h=\frac{g}{{(1+\frac{R}{R})}^2}=\frac{g}{4}.....\left(2\right)$

From eq. (1),

$\frac{T_2}{T_1}=\frac{{\sqrt{g}}_1}{{\sqrt{g}}_2}=\frac{\sqrt{g}}{{\sqrt{g}}_h}$

Now using eq.(3)

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

Hence, option $\left(d\right)$ is correct.