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 $2R$ 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 $m$
$mg=\frac{GMm}{R^2}$
where, $M$ is the mass of earth.
When it is taken to a height $h$, above the earth's surface, then
$m{g}^{\prime }=\frac{GMm}{(h+R{)}^2}$
From Eqs. (i) and (ii), we get
$g^{\prime }=g{\left(1+\frac{h}{R}\right)}^{-2}$
$=g{\left(1+\frac{2R}{R}\right)}^{-2}(\text{ given, }h=2R)$
$=g(3{)}^{-2}$
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 $2R$ above
the earth's surface is $\frac{T_1}{T_2}=\sqrt{\frac{g^{\prime }}{g}}$
$\therefore \frac{T_1}{T_2}=\sqrt{\frac{g(3{)}^{-2}}{g}}=\sqrt{\frac{1}{3^2}}$
$\Rightarrow T_2=3T_1$
$\Rightarrow \frac{T_1}{T_2}=\frac{1}{3}$