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Question

A point mass 0.1 kg in SHM of amplitude 0.1 m. When the particle passes through the mean position, its kinetic energy is 8x103J. The equation of motion of this particle if the initial phase of oscillation is 45 is:

A
y=0.1 sin (4t + π /4)
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B
y=0.1cos (4t + π/4)
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C
y=0.1 sin (2t + π /4
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D
y=0.1cos (2t + π /4)
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Solution

The correct option is B y=0.1 sin (4t + π /4)
KE=12mω2(A2x2)
x=0
8×103=12×0.1×ω2×(0.1)2
ω=4
y=Asin(ωt+π/4)
y=0.1sin(4t+π/4)

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