In S.H.M.,a simple pendulum oscillates with f | vedclass

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(C) The frequency of a simple pendulum is given by $f = \frac{1}{2\pi} \sqrt{\frac{g}{l}}$,which implies $f \propto \frac{1}{\sqrt{l}}$.Let the origin

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$f / 2$

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(C) The frequency of a simple pendulum is given by $f = \frac{1}{2\pi} \sqrt{\frac{g}{l}}$,which implies $f \propto \frac{1}{\sqrt{l}}$.Let the original length be $l_1 = l$ and the original frequency be $f_1 = f$.The length is increased by three times its original length,so the new length $l_2 = l + 3l = 4l$.Using the relation $\frac{f_2}{f_1} = \sqrt{\frac{l_1}{l_2}}$,we get $\frac{f_2}{f} = \sqrt{\frac{l}{4l}} = \sqrt{\frac{1}{4}} = \frac{1}{2}$.Therefore,the new frequency is $f_2 = \frac{f}{2}$.

In $S.H.M.$,a simple pendulum oscillates with frequency $f$. If the length of the pendulum is increased by three times its original length,then the frequency of oscillation of the pendulum will be

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