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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 energi is 18×10−3 J. The equation of motion of this particle when the initial phase of oscillation is 45o can be given by :

A
0.1cos(6t+π4)
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B
0.2sin(π4+2t)
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C
0.4sin(t+π4)
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D
0.1sin(6t+π4)
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Solution

## The correct option is D 0.1sin(6t+π4)Kinetic Energy of particle isKE=12mv2v at mean position=ωA8×10−3=120.1ω2(o.1)2ω2=16ω=4The equation isy=Asin(ωt+ϕ)y=0.1sin(4t+π4)

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