Question

A particle executing SHM of amplitude ' $a$ ' has a displacement $a/2$ at $t=T/4$ and a negative velocity. The epoch of the particle is

• A. $\frac{\pi }{3}$
• B. $\frac{2\pi }{3}$
• C. $\pi$
• D. $\frac{5\pi }{3}$

Answer 1

Answer: (A)

Equation of SHM $=y\sin (\omega t+\varphi )$
When $y=\frac{a}{2},t=\frac{T}{4}=\frac{2\pi }{4\omega }=\frac{\pi }{2\omega }$
$v=a\omega \cos (\omega t+\varphi )$,
velocity is negative
2π​<(ωt+ϕ)<23π​
$\frac{a}{2}=a\sin (\omega t+\varphi )$
$\Rightarrow \sin (\omega t+\varphi )=\frac{1}{2}$
$\left(\frac{\pi }{2}+\varphi \right)=\frac{5\pi }{6}$
Substituting in the above equation, we get $\varphi =\pi /3$.