 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 Verified
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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, ${T_{pendulum}} = 2\pi \sqrt {\dfrac{l}{g}}$ .

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.