• # question_answer 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) $\frac{\pi a\sqrt{3}}{2T}$B)  $\frac{\pi a}{T}$C) $\frac{3{{\pi }^{2}}a}{T}$D) $\frac{\pi a\sqrt{3}}{T}$

$v=\frac{dy}{dt}=A\omega \cos \omega t=A\omega \sqrt{1-{{\sin }^{2}}\omega t}$ Here, $y=\frac{a}{2}$ $\therefore$ $v=\omega \sqrt{{{a}^{2}}-\frac{{{a}^{2}}}{4}}=\omega \sqrt{\frac{3{{a}^{2}}}{4}}=\frac{2\pi }{T}\frac{a\sqrt{3}}{2}=\frac{\pi a\sqrt{3}}{T}$