A simple pendulum of length 1 m is taken to height h h

| A simple pendulum of length 1 m is taken to height h (h = R = radius of the earth) from the earth’s surface.
The time period of small oscillation of the pendulum at that point is

(g = π^{2} m s^{- 2})

A. 2 s

B. 3 s

C. 4 s

D. 1 s

Please scroll down to see the correct answer and solution guide.

Right Answer is: C

SOLUTION

At height h, $g^{/} = \frac{G M}{(R + h)^{2}} = \frac{G M}{(R + R)^{2}} = \frac{g}{4}$
$∴ T = 2 π \sqrt{\frac{l}{g^{/}}} = 2 π \sqrt{\frac{1}{\frac{1}{4} \times π^{2}}} = 2 π (\frac{2}{π}) = 4 s$

A simple pendulum of length 1 m is taken to height h h

Right Answer is: C

SOLUTION

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