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Question

# A particle of mass 400 g is executing SHM of amplitude 0.4 m. When it passes through the mean position, its kinetic energy is 32×10−3 J. If the initial phase of oscillation is π4, then the equation of motion of the particle is

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

## The correct option is C 0.4sin(t+π4) Given, Mass =400 g Amplitude (A)=0.4 m Kinetic energy at mean position (KE)=32×10−3 J Initial phase (ϕ)=π4 As we know, KE=12mω2A2 (at mean position) 32×10−3=12×0.4×ω2×(0.4)2 ω2=64×10−30.4×0.4×0.4 ⇒ω=1 rad/s ∴ Equation of motion can be written as x=Asin(ωt+ϕ) =0.4sin(t+π4)

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