Question

The amplitude and time period of a particle performing $S.H.M$ are ;a;and $T$ respectively. The displacement at which its velocity will be half the max. velocity will be

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

Answer 1

The correct option is A $\frac{a}{2}$

The displacement of a point $\times$ subjected to sumis $x=asinT$

The velocity $V$ of that point $V=acosT$

When $V=\frac{acosT}{2}$

$\Rightarrow cos\left(0.5\right)=600$

At this velocity $(V={60}^0)$ Displacement

$x=a\sin {60}^0=0.866a$