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Initial phase of an SHM that starts at half the amplitude moving in a given direction

The same question, asked 2 times: 2021 Paper 4*, 10 Apr 2023. Every wording below, oldest first.

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  1. Question 1

    2021 Paper 4

    2
     Y=Asin(ωt+ϕ0)\ Y=Asin\left( \omega t+\phi_{0} \right) is the time - displacement equation of a SHM, At t=0t= 0 the displacement of the particle is Y=A2Y=\frac{A}{2} and it is moving along negative xx-direction. Then the initial phase angle ϕ0\phi_{0} will be.

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  2. Question 2

    10 Apr 2023

    19
    A particle executes S.H.M. of amplitude A along xx-axis. At t=0t = 0, the position of the particle is x=A2x =\frac{A}{2} and it moves along positive xx-axis the displacement of particle in time tt is x=Asin(ωt+δ)x = Asin(\omega t + \delta), then the value δ\delta will be:

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