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Question

A point particle of mass 0.1kg is executing SHM with amplitude of 0.1m.When the particle passes through the mean position, its K.E is 8×103J.Obtain the equation of motion of this particle if the initial phase of oscillation is 45.

A
y=0.2sin(4t+π/2)
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B
y=0.1sin(4t+π/4)
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C
y=0.2sin(4t+π/4)
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D
None of these
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Solution

The correct option is B y=0.1sin(4t+π/4)
KE at mean is maximum.
Thus, 12mv2max=8×103 or vmax=16×102=0.4m/s
WE know that x=Asin(ωt+θ)
Thus v=dx/dt=Aωcos(ωt+θ), which has the maximum value when cosine is 1, i.e Aω
Thus, Aω=0.4
Given A=0.1, thus ω=4
Initial phase is 450=π/4
Thus B is the correct option.

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