A particle starts with S.H.M. from the mean position. Its amplitude is $A$ and its time period is $T

Question

A particle starts with S.H.M. from the mean position. Its amplitude is $A$ and its time period is $T$. At one time, its speed is half that of the maximum speed. What is its displacement?

Professor PC Thomas Classes · Answer

(d) : Maximum velocity,\ $v_{	ext {max }}=A \omega$\ According to question, $\frac{v_{\max }}{2}=\frac{A \omega}{2}=\omega \sqrt{A^{2}-y^{2}}$\ $\frac{A^{2}}{4}=A^{2}-y^{2} \Rightarrow y^{2}=A^{2}-\frac{A^{2}}{4} \Rightarrow y=\frac{\sqrt{3} A}{2}$\