Question

A simple pendulum has a time period $T_{1}$ when it is on the earth's surface and $T_{2}$ when it is taken to a height $2 R$ above the earth's surface, where $R$ is the radius of the earth. The value of $T_{1} / T_{2}$ is

• A. $1 / 9$
• B. $1 / 3$
• C. $\sqrt{3}$
• D. 9

Answer 1

Answer: (B)

As we know that, according to force of gravitation, at the surface of the earth for a simple pendulum of mass $m$
$m g=\frac{G M m}{R^{2}}$
where, $M$ is the mass of earth.
When it is taken to a height $h$, above the earth's surface, then
$m g^{\prime}=\frac{G M m}{(h+R)^{2}}$
From Eqs. (i) and (ii), we get
$g^{\prime} =g\left(1+\frac{h}{R}\right)^{-2}$
$=g\left(1+\frac{2 R}{R}\right)^{-2} (\text { given, } h=2 R)$
$=g(3)^{-2}$
The time period of a simple pendulum is given by
$T=2 \pi \sqrt{\frac{l}{g}}$
$\therefore$ Ratio of time period $T_{1}$ of a simple pendulum, when on the earth's surface and $T_{2}$ when on height $2 R$ above
the earth's surface is $\frac{T_{1}}{T_{2}}=\sqrt{\frac{g^{\prime}}{g}}$
$\therefore \frac{T_{1}}{T_{2}}=\sqrt{\frac{g(3)^{-2}}{g}}=\sqrt{\frac{1}{3^{2}}}$
$\Rightarrow T_{2}=3 T_{1}$
$ \Rightarrow \frac{T_{1}}{T_{2}}=\frac{1}{3}$