Question

The amplitude of a particle performing S.H.M. is $A$. The displacement at which its velocity will be half of the maximum velocity is

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

Answer 1

The correct option is C $\sqrt{3}A/2$

$x=Asin\omega t$
$V=\frac{dx}{dt}=A\omega cos\omega t$
$V_{max}=A\omega$
$V=\frac{V_{max}}{2}=A\omega cos\omega t$
$\Rightarrow \frac{A\omega }{2}=A\omega cos\omega t$
$\Rightarrow cos\omega t=\frac{1}{2}$
$\Rightarrow \omega t={60}^0$
$x=Asin\omega t$
$=Asin{60}^0$
$=A\frac{\sqrt{3}}{2}$
so, amplitude is $\frac{\sqrt{3}A}{2}$