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Question

The displacement of a particle from its mean position (in m) is given by y=0.2sin(10πt+1.5π)cos(10πt+0.5π). The motion of the particle is

A
Periodic but not S.H.M
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B
Non-periodic
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C
S.H.M with period 0.1 s
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D
S.H.M with period 0.2 s
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Solution

The correct option is C S.H.M with period 0.1 s
Given ,
Displacement of the particle y=0.2sin(10πt+1.5π)cos(10πt+0.5π)
From the above equation, we get
y=0.1[2sin(10πt+1.5π)cos(10πt+0.5π)]
Using, 2 sin A cos B = sin (A + B) + sin (A - B)
we get ,
y=0.1[sin(20πt+2.0π)+sin(π)]
y=0.1sin(20πt+2.0π)
y=0.1sin(20πt)
The particle starts from mean position and executes SHM.
Comparing this with y=Asin(ωt), we get ω=20π ; A=0.1 m
Time period of oscillation, T=2πω=2π20πT=110=0.1 s
Thus, option (c) is the correct answer.

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