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  4. 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 :

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Q. $Y = A \sin \left(\omega t +\phi_{0}\right)$ 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 $\phi_{0}$ will be :

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Solution:

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