Question

A particle is executing S.H.M. with amplitude $A$ and has maximum velocity $v_{0}$. Its speed at displacement $\frac{3 A}{4}$ will be

• A. $\frac{\sqrt{7}}{4} v_{0}$
• B. $\frac{v_{0}}{\sqrt{2}}$
• C. $v_{0}$
• D. $\frac{\sqrt{3}}{2} v_{0}$

Answer 1

Answer: (A)

$v=\omega \sqrt{A^{2}-x^{2}}$
$x=\frac{3}{4} A$
$v=\omega \sqrt{A^{2}-\frac{9 A^{2}}{16}}$
$v=\omega A \sqrt{\frac{7}{16}}$
or $v=v_{0} \sqrt{\frac{7}{4}}$ as $\left(v_{0}=A \omega\right)$