A simple pendulum has a time period $T_{1}$ when on the earth’s surface and $T_{2}$ when taken to a height $2R$ above the earth’s surface where $R$ is the radius of the earth. The value of $\left( \frac{T_{1}}{T_{2}} \right)$ is

Question

Dr. Abubacker · Accepted Answer

(d) The periodic time of a simple pendulum is given by,
$T = 2\pi\sqrt{\frac{l}{g}}$
When taken to height $2R$.
$g^{'} = g\left( 1 + \frac{h}{R_{e}} ight)^{2}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \  = g\left( 1 + \frac{2R}{R} ight)^{- 2} = g(3)^{- 2}$
$	herefore$ $\frac{T_{1}}{T_{2}} = \sqrt{\frac{1}{3^{2}}}$
$\Longrightarrow$ $T_{2} = 3T_{1} \Rightarrow \frac{T_{1}}{T_{2}} = \frac{1}{3}$

Q. A simple pendulum has a time period $T_{1}$ when on the earth’s surface and $T_{2}$ when taken to a height $2R$ above the earth’s surface where $R$ is the radius of the earth. The value of $\left( \frac{T_{1}}{T_{2}} \right)$ is

Explanation

Reviewed by Dr. Abubacker

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