A simple pendulum doing small oscillations at a place R height above earth surface has time period of would be it's time period if it is brought to a point which is at a height 2R from earth surface.

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Solution

Q.
A simple pendulum doing small oscillations at a place R height above earth surface has time period of $T_{1} = 4 s .$ $T_{2}$ would be it's time period if it is brought to a point which is at a height 2R from earth surface.

Choose the correct relation [R = radius of earth] [2024]

1 $2 T_{1} = 3 T_{2}$

2 $2 T_{1} = T_{2}$

3 $T_{1} = T_{2}$

4 $3 T_{1} = 2 T_{2}$

Ans.
(4)

$T = 2 π \sqrt{\frac{l}{g}} = 2 π \sqrt{\frac{l r^{2}}{GM}} = 2 π \sqrt{\frac{l}{GM}} r$

$\Rightarrow \frac{T_{2}}{T_{1}} = \frac{r_{2}}{r_{1}} = \frac{3 R}{2 R} \Rightarrow 2 T_{2} = 3 T_{1}$

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