Question

A particle is executing $S.H.M$ with amplitude $A$ and has maximum velocity $v_0$. Its speed at displacement $\frac{3A}{4}$ will be

• A. $\frac{\sqrt{7}}{4}{v}_0$
• B. $\frac{v_0}{\sqrt{2}}$
• C. $\frac{\sqrt{3}}{2}{v}_o$
• D. $v_0$

Answer 1

The correct option is A $\frac{\sqrt{7}}{4}{v}_0$

we have formula,

$v=w\sqrt{A^2-x^2}$

$v=w\sqrt{A^2-{\left(\frac{3A}{4}\right)}^2}$

$=w\sqrt{A^2-\frac{9A^2}{16}}$

$=w\sqrt{\frac{7A^2}{16}}$

$\Rightarrow v=wA\frac{\sqrt{7}}{4}$

$\therefore v=\frac{\sqrt{7}}{4}{v}_0$