A bob of simple pendulum of mass 'm' perform SHM with amplitude ' $A$ ' and period ' $T$ '. Kinetic energy…

A bob of simple pendulum of mass 'm' perform SHM with amplitude ' $A$ ' and period ' $T$ '. Kinetic energy of pendulum at displacement $\mathrm{x}=\frac{\mathrm{A}}{2}$ will be
  1. $\frac{2 m \pi^2 A}{3 T^2}$
  2. $\frac{3 \mathrm{~m} \pi^2 \mathrm{~A}}{2 \mathrm{~T}}$
  3. $\frac{2 \mathrm{~m} \pi \mathrm{A}^2}{3 \mathrm{~T}}$
  4. $\frac{3 m \pi^2 A^2}{2 T^2}$

Solution

$\begin{aligned} & \text { Kinetic energy }=\frac{1}{2} \mathrm{~m} \omega^2\left(\mathrm{~A}^2-\mathrm{x}^2\right) \\ & \text { At } \mathrm{x}=\frac{\mathrm{A}}{2}, \\ & \text { K.E. }=\frac{1}{2} \mathrm{~m} \omega^2\left(\frac{3}{4} \mathrm{~A}^2\right)=\frac{3}{8} \mathrm{~m} \omega^2 \mathrm{~A}^2 \\ & =\frac{3}{8} \mathrm{~m}\left(\frac{2 \pi}{\mathrm{T}}\right)^2 \cdot \mathrm{A}^2 \\ & =\frac{3}{8} \mathrm{~m} \frac{4 \pi^2}{\mathrm{~T}^2} \cdot \mathrm{A}^2 \\ & =\frac{3 \mathrm{~m} \pi^2 \mathrm{~A}^2}{2 \mathrm{~T}^2}\end{aligned}$

Asked in: MHT CET 2021 (24 Sep Shift 1)

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