The time period of a simple pendulum is T_1 w | vedclass

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(A) The time period of a simple pendulum is given by $T = 2\pi \sqrt{l/g}$, where $l$ is the length of the pendulum and $g$ is the acceleration due to

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$1:3$

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(A) The time period of a simple pendulum is given by $T = 2\pi \sqrt{l/g}$, where $l$ is the length of the pendulum and $g$ is the acceleration due to gravity.On the Earth's surface, $T_1 = 2\pi \sqrt{l/g}$, where $g$ is the acceleration due to gravity at the surface.At a height $h = 2R$ above the Earth's surface, the acceleration due to gravity $g_h$ is given by $g_h = g \left( \frac{R}{R+h} ight)^2$.Substituting $h = 2R$, we get $g_h = g \left( \frac{R}{R+2R} ight)^2 = g \left( \frac{R}{3R} ight)^2 = g \left( \frac{1}{3} ight)^2 = \frac{g}{9}$.The time period at height $h$ is $T_2 = 2\pi \sqrt{l/g_h} = 2\pi \sqrt{l/(g/9)} = 2\pi \sqrt{9l/g} = 3 	imes 2\pi \sqrt{l/g} = 3T_1$.Therefore, the ratio $T_1 : T_2 = T_1 : 3T_1 = 1:3$.

The time period of a simple pendulum is $T_1$ when on the Earth's surface and $T_2$ when taken to a height $h = 2R$ above the Earth's surface, where $R$ is the radius of the Earth. The ratio $T_1 : T_2$ is:

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