Question

A clock S is based on oscillation of a spring and a clock P is based on pendulum motion. Both clocks run at the same rate on earth. On a planet, having the same density as earth, but twice the radius:
(A) S will run faster than P
(B) P will run faster than S
(C) they will both run at same rates as on the earth
(D) they will both runs at equal rates, but not the same as on earth

Answer 1

Hint : Oxidation defines the to and fro motion of body. Pendulum is a body suspended from a fixed point so as to swing freely to and fro under the action of quality. A spring is a device that stores potential energy, specifically elastic potential energy.

Formula:
Time period of spring, $ {T_{spring}} $ $ = 2\pi \sqrt {\dfrac{m}{k}} $
Time period of pendulum, $ {T_{pendulum}} = 2\pi \sqrt {\dfrac{l}{g}} $ .

Complete step by step answer
According to the question, clock S and clock P run at the same rate on earth.
We know that clock S is based on the oxidation of spring.
So, time period of spring which is clock S $ {T_s} = 2\pi \sqrt {\dfrac{m}{k}} $
Time period of clock S depends on mass of body $ \left( m \right) $ and spring constant $ \left( k \right) $
Also, the time period of clock P which is based on pendulum motion is given $ {T_P} = 2\pi \sqrt {\dfrac{l}{g}} $
Time period of clock P depends on length of pendulum $ \left( l \right) $ and acceleration due to gravity $ \left( g \right) $
We know that, $ g = \dfrac{{GM}}{{{R^2}}} $ , R $ = $ Radius of earth or planet
So, g $ \propto $ $ \dfrac{1}{{{{\left( {2R} \right)}^2}}} $ $ = > $ g $ \propto \dfrac{1}{{4{R^2}}} $
As radius is doubled, g decreases so the time period of clock P also decreases as it depends on R, which is different on different planets.
But, time period of clock S will remain same as $ {T_s} = 2\pi \sqrt {\dfrac{m}{k}} $
Which depends on mass but not on weight, So time period of S will be the same on other planets which are double the radius of earth.
So, S will run faster than P
Option $ \left( A \right) $ is correct.

Note
A clock runs slow, when the time period of its pendulum is increased and runs fast, when time period is decreased.