The period of a simple pendulum gets doubled when

The period of a simple pendulum gets doubled when
  1. its length is doubled.
  2. its length is made four times.
  3. its length is made half.
  4. the mass of the bob is doubled.

Solution

We know for a simple pendulum, $\begin{aligned} & \mathrm{T}=2 \pi \sqrt{\frac{l}{\mathrm{~g}}} \Rightarrow \mathrm{~T} \propto \sqrt{l} \\ & \frac{\mathrm{~T}_1}{\mathrm{~T}_2}=\sqrt{\frac{l_1}{l_2}} \end{aligned}$ When $l_2=4 l_1$, $\frac{\mathrm{T}_1}{\mathrm{~T}_2}=\sqrt{\frac{l_1}{4 l_1}} \Rightarrow \mathrm{~T}_2=2 \mathrm{~T}_1$ $\therefore \quad$ Period is doubled when length is made four times $\begin{aligned} & \frac{\mathrm{T}_1}{2 \mathrm{~T}_1}=\sqrt{\frac{l_1}{l_2}} \\ & \frac{1}{2}=\sqrt{\frac{l_1}{l_2}} \Rightarrow \frac{1}{4}=\frac{l_1}{l_2} \Rightarrow l_2=4 l_1 \end{aligned}$ $\therefore \quad$ Period is doubled when length is made four times

Asked in: MHT CET 2024 (03 May Shift 2)

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