First slide

Variation of g

Question

Time period of oscillation of a simple pendulum on the surface of earth is T. At what height above the surface of earth will the time period be 4T ?

Moderate

A

$\frac{3 R}{2}$
B

R
C

3R
D

2R

Solution

$T = 2 π \sqrt{\frac{l}{g}} and 4 T = 2 π \sqrt{\frac{l}{g^{'}}} ∴ \frac{T}{4 T} = \sqrt{\frac{g^{'}}{g}} = \frac{R}{R + h} \Rightarrow h = 3 R$

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