Question

A simple pendulum of length 4 m is taken to a height $R$ (radius of the earth) from the earth's surface.The time period of small oscillations of the pendulum is $(g_{surface}={\pi }^{2}m{s}^{-2})$

• A. 2 s
• B. 4 s
• C. 8 s
• D. 16 s

Answer 1

The correct option is C 8 s

The acceleration due to gravity of earth at a high 'h'

$g_h=g_{surface}{(1+\frac{h}{R})}^{-2}$

g = acceleration due to surface gravity at earth surface

Now h = R (given)

so, $g_h=\frac{g_{surface}}{4}$

Now the time period of simple pendulum of length 4m at earth surface is

$T_{surface}=2\pi \sqrt{\frac{l}{g_{surface}}}=2\pi \sqrt{\frac{4}{9.8}}\approx 4sec$

So time period at hight 'h = 2R' is

$T=2\pi \sqrt{\frac{l}{g_h}}=2\pi \sqrt{\frac{4\times 4}{g_{surface}}}=8sec$