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. $2$
• D. $4$

Answer 1

The correct option is C $2$

The time period of a simple pendulum depends upon the acceleration due to gravity and is given by$$T=2\pi \sqrt{\frac{l}{g}}$$ $\Rightarrow T\propto \frac{1}{\sqrt{g}}...\left(1\right)$

Acceleration due to gravity at a height $h$ above the surface of the Earth,$$g_h=\frac{GM}{(R+h{)}^2}...\left(2\right)$$ But acceleration due to gravity at the surface of the Earth, $$g=\frac{GM}{R^2}...\left(3\right)$$ From equation $\left(2\right)$ and $\left(3\right)$, we have

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

At $h=R$,

$\Rightarrow g^{\prime }=g{\left(\frac{R}{R+R}\right)}^2$

$\Rightarrow \frac{g^{\prime }}{g}=\frac{1}{4}$

Using equation $\left(1\right)$,

$\frac{T_2}{T_1}=\sqrt{\frac{g}{g^{\prime }}}=\sqrt{\frac{4}{1}}$

$\therefore \frac{T_2}{T_1}=2$

Hence, option (d) is the correct answer.