Q106 / 150 · MHT CET 2023 (12 May Shift 1) · Physics

A simple pendulum performs simple harmonic motion about $\mathrm{x}=0$ with an amplitude ' $\mathrm{a}$ ' and time period ' $T$ '. The speed of the pendulum at $x=\frac{a}{2}$ is

  • A
    $\frac{\pi \mathrm{a}}{\mathrm{T}}$
  • B
    $\frac{3 \pi^2 \mathrm{a}}{\mathrm{T}}$
  • Correct answer
    C
    $\frac{\pi \mathrm{a} \sqrt{3}}{\mathrm{~T}}$
  • D
    $\frac{\pi \mathrm{a} \sqrt{3}}{2}$

Solution

$\begin{aligned} \mathrm{v} & =\omega \sqrt{\mathrm{a}^2-\mathrm{x}^2} \\ \text { At } \mathrm{x} & =\frac{\mathrm{a}}{2}, \\ \therefore \quad \mathrm{v} & =\omega \sqrt{\mathrm{a}^2-\frac{\mathrm{a}^2}{4}} \\ & =\omega \frac{\sqrt{3 \mathrm{a}}}{2} \\ & =\frac{2 \pi}{\mathrm{T}} \times \frac{\sqrt{3} \mathrm{a}}{2} \\ \therefore \quad \mathrm{v} & =\frac{\pi \mathrm{a} \sqrt{3}}{\mathrm{~T}}\end{aligned}$

Hence, the correct answer is option: 'C'.

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