Question

1. The displacement of a particle is represented by the
equation y = 3 cos (π/4-2wt). The motion of the particle is

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Answer 1

Answer:

Explanation:

The displacement of the particle

y=3cos(π4−2ωt) velocity of the particle

v=dydt=ddt[3cos(π4−2ωt)]

=6ωsin(π4−2ωt)

Acceleration

a=dvdt=ddt[6ωsin(π4−2ωt)]

=−12ω2cos(π4−2ωt)=

−4ω2[3cos(π4−2ωt)]

Here a=−4ω2y=−(2ω)2y

It means acceleration, aα−y, the motions is SHM. Hence angular frequency of S.H.M,ω'=2ω

ω'=2ω=2πT'⇒T'=2π2ω=πω

It means the motion is SHM with period πω.

METHOD2: Given the equation of displacement of the particle

y=3cos(π4−2ωt)

y=3cos[−(2ωt−π4)]

We know cos(−θ)=cosθ

Hence y=3cos(2ωt−π4)

Compering with y=acos(ωt+ϕ0)

Hence (i) reprecents simple hamronic motion with angular frequency 2ω.

Hence its time period, T=2π2ω=πω.