Question

If ${T_1}$ , ${T_2}$ and ${T_3}$ are the time periods of a given simple pendulum on the surface of the earth, at a depth of $h$ in a mine and at an altitude $h$ above the earth’s surface respectively, then
(A) ${T_1} = {T_2} = {T_3}$
(B) ${T_2} < {T_1} > {T_3}$
(C) ${T_2} > {T_1} < {T_3}$
(D) ${T_1} > {T_2} > {T_3}$

Answer 1

Hint
The time period of oscillation of a pendulum is given by the formula, $T = 2\pi \sqrt {\dfrac{l}{g}}$ . So when the value of $g$ changes with the altitude or the depth, the time period of oscillation changes accordingly. Therefore, the time period of oscillation changes with change in $g$ .
Formula used: $T = 2\pi \sqrt {\dfrac{l}{g}}$
where $T$ is the time period of oscillation of the pendulum, $l$ is the length of the pendulum and $g$ is the acceleration due to gravity.
${g_h} = g{\left( {\dfrac{R}{{R + h}}} \right)^2}$ where ${g_h}$ is the acceleration due to gravity at height $h$ , $R$ is the radius of earth and $h$ is the height above the earth’s surface.
and ${g_d} = g\left( {1 - \dfrac{d}{R}} \right)$ where ${g_d}$ is the acceleration due to gravity at depth $d$ .

Complete step by step answer
The acceleration due to gravity is the acceleration that is gained by a freely falling object due to the gravitational force on earth. It is denoted by the symbol $g$ and has a value $9.8m/{s^2}$ . The value of acceleration due to gravity is constant all over the surface of the earth but changes with the increase in altitude over the earth’s surface or the increase in depth below the earth’s surface. In the question, the time period of a pendulum is given ${T_1}$ , ${T_2}$ and ${T_3}$ on the surface of the earth, at a depth of $d$ in a mine and an altitude $h$ above the earth’s surface respectively.
For a pendulum on the earth’s surface, the time period is given by ${T_1} = 2\pi \sqrt {\dfrac{l}{g}}$ .
Therefore, ${T_1} \propto \dfrac{1}{{\sqrt g }}$ .
The change in the value of $g$ due to depth is given by ${g_d} = g\left( {1 - \dfrac{d}{R}} \right)$ .
So, as the depth below the earth’s surface increases, the value of $d$ increases. Since the value $\left( {1 - \dfrac{d}{R}} \right)$ is always less than 1, by multiplying $g$ with a value less than 1, the value ${g_d}$ becomes less than $g$ .
Hence, the time period of the pendulum at depth $d$ is
${T_2} = 2\pi \sqrt {\dfrac{l}{{{g_d}}}}$
$\therefore {T_2} \propto \dfrac{1}{{\sqrt {{g_d}} }}$
Since ${g_d} < g$ ,
$\therefore \dfrac{1}{{{g_d}}} > \dfrac{1}{g}$
So we get $\therefore {T_2} > {T_1}$
The change in the value of $g$ due to altitude is given by ${g_h} = g{\left( {\dfrac{R}{{R + h}}} \right)^2}$ .
As the altitude above the earth’s surface increases, the value of $h$ increases. Therefore, the value of the fraction $\dfrac{R}{{R + h}}$ becomes less than 1 as the value of the denominator in more than the numerator. So, by multiplying $g$ with a value less than 1, the value of ${g_h}$ becomes less than $g$ .
Hence, the time period of the pendulum at an altitude $h$ is
${T_3} = 2\pi \sqrt {\dfrac{l}{{{g_h}}}}$
$\therefore {T_3} \propto \dfrac{1}{{\sqrt {{g_h}} }}$
Since ${g_h} < g$ ,
$\therefore \dfrac{1}{{{g_h}}} > \dfrac{1}{g}$
So, we get $\therefore {T_3} > {T_1}$
Therefore, we have obtained, ${T_2} > {T_1}$ and ${T_3} > {T_1}$ . Combining we get, ${T_2} > {T_1} < {T_3}$.
Hence, option (C) is correct.

Note
As we increase the altitude above the earth’s surface the value of acceleration due to gravity decreases because the distance from the center of the earth increases. And as we do down a depth from the surface of the earth, then the total mass enclosed within that radius decreases and so the gravity decreases.