If the mass of the bob in a simple pendulum is increased to thrice its original mass and its length is made half its original length, then the new time period of oscillation is $\frac{x}{2}$ times its original time period. Then the value of $x$ is:

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Aicharya · Accepted Answer

$\begin{aligned} & T^{\prime}=2 \pi \sqrt{\frac{\ell^{\prime}}{g}} 	ext { where } \ell^{\prime}=\frac{\ell}{2} \ & T=2 \pi \sqrt{\frac{\ell}{g}} \ & T^{\prime}=\frac{x}{2} T \ & 2 \pi \sqrt{\frac{\ell}{2 g}}=\frac{x}{2} 2 \pi \sqrt{\frac{\ell}{g}} \ & \frac{1}{\sqrt{2}}=\frac{x}{2} \Rightarrow x=\sqrt{2}\end{aligned}$

If the mass of the bob in a simple pendulum is increased to thrice its original mass and its length is made…

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