Time period of a simple pendulum on earth's surface is 'T'. It time period becomes 'xT' when taken to a height '2R' above earth's surface. The value of x will b

Question

Time period of a simple pendulum on earth's surface is 'T'. It time period becomes 'xT' when taken to a height '2R' above earth's surface. The value of x will be ( R = radius of earth)

TestHub · Accepted Answer

Time period of a simple pendulum on the earth's surface is given by:

$T = 2 π \frac{L}{g}$

Acceleration due to gravity at a height $h$ above the earth's surface is:

$g_{h} = g (\frac{R}{R + h})^{2}$

For $h = 2 R$ :

$g_{h} = g (\frac{R}{R + 2 R})^{2} = g (\frac{1}{3})^{2} = \frac{g}{9}$

Time period at height $h = 2 R$ is:

$T_{h} = 2 π \frac{L}{g _{h}} = 2 π \frac{L}{g /9} = 3 \times 2 π \frac{L}{g} = 3 T$

Given $T_{h} = x T$ , we get $x = 3$ .

Answer: $3$

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