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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 8×10−3J.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=8×10−3 or vmax=√16×10−2=0.4m/sWE know that x=Asin(ωt+θ)Thus v=dx/dt=Aωcos(ωt+θ), which has the maximum value when cosine is 1, i.e AωThus, Aω=0.4Given A=0.1, thus ω=4Initial phase is 450=π/4Thus B is the correct option.

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