The Time period of oscillation of a simple pendulum whose length is very large (>> Radius of earth R) is T. What will be the time period of a simple pendulum whose length is equal to the radius of earth(R) ?

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$\frac{T}{2 \sqrt{2}}$

$\frac{T}{2}$

$\sqrt{2} T$

$\frac{T}{\sqrt{2}}$

answer is D.

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Detailed Solution

For a pendulum whose length( $l$ ) is comparable to the radius of earth(R), time period is given by

$T = \frac{2 π}{\sqrt{g (\frac{1}{l} + \frac{1}{R})}} If l > > R, T = 2 π \sqrt{\frac{R}{g}} If l = R, T^{'} = 2 π \sqrt{\frac{R}{2 g}} ∴ \frac{T^{'}}{T} = \frac{1}{\sqrt{2}} \Rightarrow T^{'} = \frac{T}{\sqrt{2}}$

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