A simple pendulum has a time period {T₁} when on earth’s surface and {T₂} when taken to a

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7.Gravitation

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A simple pendulum has a time period ${T_1}$ when on the earth’s surface and ${T_2}$ when taken to a height $R$ above the earth’s surface, where $R$ is the radius of the earth. The value of ${T_2}/{T_1}$ is

A

$1$

B

$\sqrt 2 $

C

$4$

D

$2$

(IIT-2001)

Solution

(d) If acceleration due to gravity is $g$ at the surface of earth then at height $R$ it value becomes

$g' = g{\left( {\frac{R}{{R + h}}} \right)^2} = \frac{g}{4}$

${T_1} = 2\pi \sqrt {\frac{l}{g}} {\rm{ and }}\,{T_2} = 2\pi \sqrt {\frac{l}{{g/4}}} $

$\frac{{{T_2}}}{{{T_1}}} = 2$

Standard 11

Physics

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