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Question

The initial conditions of a particle executing SHM are as shown in the figure. Find the phase constant ϕ of the particle.
Given: Range of ϕ is [0,2π]


A
7π4
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B
π4
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C
5π4
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D
3π4
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Solution

The correct option is C 5π4
Equation of a particle executing SHM, is given by x=Asin(ωt+ϕ) ......(1)
where ϕ is in the range of [0,2π]
From the given figure, we can deduce that, at t=0x=A2
Now from equation (1), we can write that
A2=Asinϕsinϕ=12ϕ=5π4 or 7π4
Differentiating (1) with respect to time on both sides, we get
v=Aωcos(ωt+ϕ)
Also, from the figure, we can deduce that the particle is moving towards negative extreme position. So, we can conclude that at t=0,
v=Aωcosϕ<0cosϕ<0ϕ=5π4
Thus, option (c) is the correct answer.

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