Question

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

• A. $\frac{2A}{\sqrt{3}}$
• B. $\frac{3A}{\sqrt{2}}$
• C. $\frac{\sqrt{2}A}{3}$
• D. $\frac{\sqrt{3}A}{2}$

Answer 1

Answer: (D)

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