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

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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:

Accepted Answer

Sol. [Key Concept: Simple pendulum — period independent of bob mass] $T^{'} = 2\pi\sqrt{\frac{\mathcal{l}^{'}}{g}}$ where $\mathcal{l}^{'} = \frac{\mathcal{l}}{2}$$$\begin{matrix} & T = 2\pi\sqrt{\frac{\mathcal{l}}{g}} \ & T^{'} = \frac{x}{2}T \ & 2\pi\sqrt{\frac{\mathcal{l}}{2g}} = \frac{x}{2}2\pi\sqrt{\frac{\mathcal{l}}{g}} \ & \frac{1}{\sqrt{2}} = \frac{x}{2} \Rightarrow x = \sqrt{2} \end{matrix}$$

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