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 \[x=\frac{a}{2}\] will be                                                                   [AIPMT (S) 2009]

A)  \[\frac{\pi a\sqrt{3}}{2T}\]        

B)       \[\frac{\pi a}{T}\]                     

C)  \[\frac{3{{\pi }^{2}}a}{T}\]   

D)       \[\frac{\pi a\sqrt{3}}{T}\]

Correct Answer: A

Solution :

\[v=\frac{dy}{dt}=A\,\omega \,\cos \,\,\omega t\,=\,A\,\omega \,\sqrt{1-{{\sin }^{2}}\,\omega \,t}\]

\[=\omega \,\sqrt{{{A}^{2}}-{{y}^{2}}}\]

Here,     \[y=\frac{a}{2}\]

\[\therefore \]