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Question

# A point particle of mass 0.1 kg is executing S.H.M. of amplitude of 0.1 m. When the particle passes through the mean position, its kinetic energy is 8×10−3Joule. Obtain the equation of motion of this particle if this initial phase of oscillation is 45∘.

A
y=0.1sin(±4t+π4)
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B
y=0.2sin(±4t+π4)
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C
y=0.1sin(±2t+π4)
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D
y=0.2sin(±2t+π4)
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Solution

## The correct option is A y=0.1sin(±4t+π4)The displacement of a particle in S.H.M. is given by:y=asin(ωt+ϕ)velocity=dydt=ωacos(ωt+ϕ)The velocity is maximum when the particle passes through the mean position i.e.(dydt)max=ωaThe kinetic energy at this instant is given by12m(dydt)2max=12mω2a2=8×10−3jouleor 12×(0.1)ω2×(0.1)2=8×10−3Solving we get ω=±4Substituting the values of a,ω and ϕ in the equation of S.H.M., we gety=0.1sin(±t+π/4)metre.

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