The ratio of the time periods of a simple pen | vedclass

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(C) The time period of a simple pendulum is given by $T = 2\pi \sqrt{\frac{L}{g'}}$.Here,$g'$ is the acceleration due to gravity at a height $h$ from

Vedclass · Accepted Answer

$3: 4$

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(C) The time period of a simple pendulum is given by $T = 2\pi \sqrt{\frac{L}{g'}}$.Here,$g'$ is the acceleration due to gravity at a height $h$ from the surface of the earth,given by $g' = g  (\frac{R_E}{R_E + h} )^2$.Therefore,$T \propto \frac{1}{\sqrt{g'}} \propto \frac{R_E + h}{R_E}$.At height $h_1 = 2 R_E$,the time period is $T_1 \propto \frac{R_E + 2 R_E}{R_E} = \frac{3 R_E}{R_E} = 3$.At height $h_2 = 3 R_E$,the time period is $T_2 \propto \frac{R_E + 3 R_E}{R_E} = \frac{4 R_E}{R_E} = 4$.The ratio of the time periods is $\frac{T_1}{T_2} = \frac{3}{4}$ or $3: 4$.

The ratio of the time periods of a simple pendulum at heights $2 R_E$ and $3 R_E$ from the surface of the earth is ($R_E$ is the radius of the earth).

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