Question

The time period of oscillation of a particle that executes S.H.M. is $1.2sec$. The time starting from extreme position, its velocity will be half of its velocity at mean position is

• A. $0.1$ s
• B. $0.2$ s
• C. $0.4$ s
• D. $0.6$ s

Answer 1

The correct option is A $0.1$ s

Time is zero at extreme position

$T=1.2sec$

$v=\omega \sqrt{A^2-x^2};v_{max}=\omega A$

$v=\frac{v_{max}}{2}=\frac{\omega A}{2}=\omega$ $\sqrt{A^2-x^2}$

$\frac{A^2}{4}=A^2-x^2;x=\frac{\sqrt{3}}{2}A$

$x=A$ cos $\omega t$

$\frac{\sqrt{3}}{2}A=A$ cos $\omega t$

$\omega t=\frac{\pi }{6}$

$t=\frac{\pi }{6\times \omega }$

$t=\frac{\text{/}\pi }{6}\times \frac{1.2}{2\mathrm{\pi /}}=0.1sec$