Y=A sin(ωt+ϕ0) is the time –displacement equation of a SHM. At t=0 the displacement of the particle is Y=A 2 and it is moving along negative x-direction. Then the initial phase angle ϕ0 will be:

y=Asin(ωt+ϕo), at t=0: y=−A/2, −A2=Asinϕo, ϕo=5π6velocity is positive for ϕo=π6 velocity is negaitive for ϕo=5π6

y=Asin(ωt+ϕo), at t=0: y=−A/2, −A2=Asinϕo, ϕo=5π6velocity is positive for ϕo=π6 velocity is negaitive for ϕo=5π6

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$Y = A \sin (ω t + ϕ_{0})$ is the time –displacement equation of a SHM. At t=0 the displacement of the particle is $Y = \frac{A}{2}$ and it is moving along negative x-direction. Then the initial phase angle $ϕ_{0}$ will be:

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$\frac{π}{3}$

$\frac{π}{6}$

$\frac{5 π}{6}$

$\frac{2 π}{3}$

answer is D.

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$y = A \sin (ω t + ϕ_{o}), a t t = 0 : y = - A / 2, - \frac{A}{2} = A \sin ϕ_{o}, ϕ_{o} = 5 \frac{π}{6}$

$v e l o c i t y i s p o s i t i v e f o r ϕ_{o} = \frac{π}{6} v e l o c i t y i s n e g a i t i v e f o r ϕ_{o} = \frac{5 π}{6}$

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