Question

The time period of oscillation of a particle, that executes SHM, is $1.2s$. The time, starting from mean position, at which its velocity will be half of its velocity at mean position is ?

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

Answer 1

The correct option is A $0.1s$

Time is zero at extreme position

$T=1.2sec$

$u=\omega \sqrt{A^2-x^2};u_{max}=\omega A$

$u=\frac{u_{max}}{2}=\frac{\omega A}{2}=\omega \sqrt{A^2-x^2}$

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

$x=Acos\omega t$

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

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

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

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