Q. A particle starts S.H.M. from the mean position. Its amplitude is $A$ and time period is $T.$ At the time when its speed is half of the maximum speed, its displacement $y$ is
A) $\frac{A}{2}$
B) $\frac{A}{\sqrt{2}}$
C) $\frac{A\sqrt{3}}{2}$
D) $\frac{2A}{\sqrt{3}}$
Explanation
(c) $v_{\max} = \omega A \Rightarrow v = \frac{\omega A}{2} = \omega\sqrt{A^{2} - y^{2}}$
$\Rightarrow A^{2} - y^{2} = \frac{A^{2}}{4} \Rightarrow y^{2} = \frac{3A^{2}}{4} \Rightarrow y = \frac{\sqrt{3}A}{2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.