Y = A sin (ω t +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 :

Question

Y = A sin (ω t +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 : (A) (π/6) (B) (π/3) (C) (5 π/6) (D) (2 π/3)

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/43bc95cebea565e14de23deef8dfc3b0-.png" /> initial phase (π/2)+(π/3)=(5 π/6)

Q. $Y = A sin (ω t + ϕ_{0})$ 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 :

$\frac{π}{6}$

$\frac{π}{3}$

$\frac{5 π}{6}$

$\frac{2 π}{3}$

initial phase $\frac{π}{2} + \frac{π}{3} = \frac{5 π}{6}$

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

6π​

3π​

65π​

32π​