Question

A high altitude experiment was conducted where it was found that $10\% $ of the length of the pendulum had to be cut to preserve the time period as it was on the Earth’s surface. What is the height (in terms of radius of the earth $R$) at which the experiment was done?
A. $\dfrac{{9R}}{{20}}$
B. $\dfrac{R}{{10}}$
C. $1.50R$
D. $0.05R$

Answer 1

Hint: A pendulum is a weight that is suspended from a hinge and can freely swing. When a pendulum is displaced sideways from its resting, equilibrium position, gravity acts as a restoring force, accelerating the pendulum back to its equilibrium position.

Complete step by step answer:
A Pendulum is a body that is suspended from a fixed point and swings back and forth under the force of gravity. Pendulums are used to control the movement of clocks since the duration, or time interval, for each complete oscillation is constant.
Time period of pendulum = $T = 2\pi \sqrt {\dfrac{l}{g}} $
At height,$l$ is changed to $l'$
$l' = l - \dfrac{{10l}}{{100}} = \dfrac{{9l}}{{10}}$
$T' = T$ and $g$ at this height = $g'$
$\Rightarrow g'= \dfrac{{GM}}{{\left( {R + h} \right)}}$
$\Rightarrow g' = \dfrac{{GM}}{{{R^2}{{\left( {1 + \dfrac{h}{R}} \right)}^2}}} \\
\Rightarrow g'= \dfrac{g}{{\left( {1 + \dfrac{h}{R}} \right)}} \\ $
Now,
$T' = T$
$ \Rightarrow 2\pi \sqrt {\dfrac{{l'}}{{g'}}} $= $2\pi \sqrt {\dfrac{l}{g}} $
$ \Rightarrow \sqrt {\dfrac{{l'}}{{g'}}} = \sqrt {\dfrac{l}{g}} $
$ \Rightarrow \sqrt {\dfrac{{9l}}{{10}}\dfrac{{\left( {1 + \dfrac{h}{R}} \right)}}{g}} = \sqrt {\dfrac{l}{g}} $
$ \Rightarrow \sqrt {\dfrac{{9l}}{{10}}} \left( {1 + \dfrac{h}{R}} \right) = 1$
$ \Rightarrow 1 + \dfrac{h}{R} = \dfrac{{\sqrt {10} }}{3}$
$ \Rightarrow h = \left( {\dfrac{{\sqrt {10} }}{3} - 1} \right)R$
$\therefore h = 0.05R$
Therefore, the height (in terms of radius of the earth R) at which the experiment was done is $0.05R$

Hence, correct option is D.

Note: The pendulum's duration is affected by the length of the string; the longer the string, the longer the pendulum's period. A pendulum with a longer string has a lower frequency, which means it swings back and forth less often in a given period of time than one with a shorter string.