A simple pendulum of length $l_1$ has time period $T_1$. Another simple pendulum of length $l_2\left(l_1 \gt…

A simple pendulum of length $l_1$ has time period $T_1$. Another simple pendulum of length $l_2\left(l_1 \gt l_2\right)$ has time period $T_2$. Then the time period of the pendulum of length $\left(l_1-l_2\right)$ will be
  1. $T_1-T_2$
  2. $\sqrt{\frac{T_1}{T_2}}$
  3. $\sqrt{\mathrm{T}_1^2-\mathrm{T}_2^2}$
  4. $\sqrt{\frac{T_2}{T_1}}$

Solution

For a simple pendulum, $\mathrm{T}_1=2 \pi \sqrt{\frac{l_1}{\mathrm{~g}}} \text { and } \mathrm{T}_2=2 \pi \sqrt{\frac{l_2}{\mathrm{~g}}}$
Now, $\mathrm{T}_1{ }^2=4 \pi \times \frac{l_1}{\mathrm{~g}}$ and $\mathrm{T}_2^2=4 \pi \times \frac{l_2}{\mathrm{~g}}$ $\therefore \quad l_1=\frac{\mathrm{T}_1^2 \mathrm{~g}}{4 \pi}$ and $l_2=\frac{\mathrm{T}_2^2 \mathrm{~g}}{4 \pi}$ $\therefore \quad l_1-l_2=\left(\mathrm{T}_1^2-\mathrm{T}_2^2\right) \times \frac{\mathrm{g}}{4 \pi}$ Time period of pendulum of length $\left(l_1-l_2\right)$ is, $\begin{aligned} & \mathrm{T}=2 \pi \sqrt{\frac{\left(l_1-l_2\right)}{\mathrm{g}}}=2 \pi \sqrt{\frac{\left(\mathrm{~T}_1^2-\mathrm{T}_2^2\right) \times \frac{\mathrm{g}}{4 \pi}}{\mathrm{~g}}} \\ & \therefore \quad \mathrm{~T}=\sqrt{\mathrm{T}_1^2-\mathrm{T}_2^2} \end{aligned}$ /

Asked in: MHT CET 2024 (04 May Shift 1)

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