Q123 / 180 · AIIMS 2013 · Physics

A simple pendulum performs simple harmonic motion about $x=0$ with an amplitude ' $a$ ' and time period ' $T$ '. The speed of the pendulum at $x=a / 2$ will be

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

Solution

As simple pendulum performs simple harmonic motion. $\therefore$ velocity, $v=\omega \sqrt{a^2-x^2}$ At, $x=\frac{a}{2}$ $v=\frac{2 \pi}{T} \sqrt{a^2-\left(\frac{a}{2}\right)^2}=\frac{2 \pi}{T} \frac{\sqrt{3 a^2}}{2}=\frac{\pi a \sqrt{3}}{T}$

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

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