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Question

The motion of a particle executing simple harmonic motion is described by the displacement function x(t)=Acos(wt+ϕ). If the initial ( t= 0) position of the particle is 1 cm and its initial velocity is w 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(wt+α) , what are the amplitude and initial phase of particle with the above initial conditions?

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Solution

intiatlly at t=0,
displacement x=1cm
initial velocity, v=ωcm/sec.
angular frequency, ω=πrad/s1
its is given that,
x(t)=Acos(wt+ϕ)
1=Acos(wt×0=ϕ)=Acos ϕ
Acos ϕ=1 . . . . . . . . .(1)


now velocity, v=dxdt
ω=Aωsin(ωt+ϕ)
Asinϕ=1 . . . . . . . . .(2)

squaring and adding equations (1)and (2) we get,
A2(sin2ϕ+cos2ϕ)=1+1
A2(sin2ϕ+cos2ϕ)1+1
A=2 cm

dividing equation (2) and (1) we get
tanϕ=1
ϕ3π4,7π4,......

SHM is given as :
x=Bsin(wt+α)
putting the given values in this equation, we get:
1=Bsin[ω×0+α]=1+1
Bsinα=1 . . . . . . . . . .(3)

velocity, v=ωBcos(wt+α)
substituting the given values, we get :
B2[sin2α+cos2α]=1=1
Bsinα=1 . . . . . . . . . . .(4)

squaring and adding equation 3 and 4 we get
B2[sin2α+cos2α]=1+1
B=2 cm

Dividing equation (3) by (4) we get,
BsinαBcosα=11
tanα=1=tanπ4

α=π4,5π4

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