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Question

The displacement of a particle from its mean position (in metres) is given by y=0.2sin(10πt+1.5π)cos(10πt+1.5π). The motion of the particle is SHM, with a time period of (in seconds)

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Solution

Given, displacement
y=0.2sin(10πt+1.5π)cos(10πt+1.5π)
Considering (10πt+1.5π)=θ
y=0.2sinθcosθ
Multiplying and dividing by 2,
y=0.1×sin2θ (2sinθcosθ=sin2θ)
Putting value of θ back,
y=0.1×sin2(10πt+1.5π)
=0.1sin(20πt+3π)

Comparing with standard equation of SHM y=Asin(ωt+ϕ),
we get, ω=20π
Time period is given as T=2π/ω
T=2π20π
T=0.1 s
Hence, motion of the particle is simple harmonic with time period =0.1 s

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