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Question

The motion of a particle executing simple harmonic motion is described by the displacement function, x(t)=Acos(ωt+ϕ).
If the initial (t=0) position of the particle is 1 cm and its initial velocity is ω cm/s, what are its amplitude and initial phase angle ? The angular frequency of the particle is πs1. If instead of the cosine function we choose the sine function to describe the SHM:x=Bsin(ωt+α), what are the amplitude and initial phase of the particle with the above initial conditions.

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Solution

Intially, at t = 0;
Displacement, x=1cm
Intial velocity, v=ωcm/sec.
Angular frequency, ω=πrad/s1
It is given that,
x(t)=Acos(ωt+ϕ)
1=Acos(ω×0+ϕ)=Acosϕ
Acosϕ=1 ...(i)
Velocity, v=dx/dt
ω=Aωsin(ωt+ϕ)
1=Asin(ω×0+ϕ)=Asinϕ
Asinϕ=1 ...(ii)

Squaring and adding equations (i) and (ii), we get:
A2(sin2ϕ+cos2ϕ)=1+1
A2=2
A=2cm
Dividing equation (ii) by equation (i), we get:
tanϕ=1
ϕ=3π/4,7π/4
SHM is given as:
x=Bsin(ωt+α)
Putting the given values in this equation, we get:
1=Bsin(ω×0+α]=1+1
Bsinα=1 ...(iii)
Velocity, v=ωcos(ωt+α)
Substituting the given values, we get:
π=πBcosα
Bcosα=1 ...(iv)
Squaring and adding equations(iii) and (iv), we get:
B2[sin2α+cos2α]=1+1
B2=2
B=2
Dividing equation (iii) by equation (iv), we get:
Bsinα/Bcosα=1/1
tanα=1=tanπ/4
α=π/4,5π/4,......

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