Q112 / 197 · NEET 2009 (Mains) · Physics

A simple pendulum performs simple harmonic motion about $\mathrm{x}=0$ with an amplitude a and time period T. The speed of the pendulum at $\mathrm{x}=\mathrm{a} / 2$ will be :

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

Solution

$\quad \mathrm{v}=\omega \sqrt{\mathrm{A}^2-\mathrm{x}^2}$ $$ =\frac{2 \pi}{\mathrm{T}} \sqrt{\mathrm{a}^2-\frac{\mathrm{a}^2}{4}}=\frac{\pi \mathrm{a} \sqrt{3}}{\mathrm{~T}} $$

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

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