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Question

The motion of a mass on a spring, with spring constant K is as shown in figure.


The equation of motion is given by x(t)=Asinωt+Bcosωt with ω=Km.

Suppose that at time t=0, the position of mass is x(0) and velocity v(0), then its displacement can also be represented as x(t)=Ccos(ωtϕ), where C and ϕ are :

A
C=v(0)2ω2+x(0)2, ϕ=tan1(v(0)x(0)ω)
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B
C=2v(0)2ω2+x(0)2, ϕ=tan1(x(0)ω2v(0))
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C
C=2v(0)2ω2+x(0)2, ϕ=tan1(x(0)ωv(0))
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D
C=2v(0)2ω2+x(0)2, ϕ=tan1(v(0)x(0)ω)
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Solution

The correct option is A C=v(0)2ω2+x(0)2, ϕ=tan1(v(0)x(0)ω)
Given,

x(t)=Asinωt+Bcosω t

v=dxdt=AωcosωtBωsinωt

At t=0,x(0)=B and v(0)=Aω

x=A sinωt+B sin (ωt+90)

Now,


Net amplitude, Anet=A2+B2

From figure, tanα=BAcotα=AB

Now, the equation can be written as

x=A2+B2sin(ωt+α)

x=A2+B2cos(ωt(90α))

Given, x=Ccos(ωtϕ)

C=A2+B2

Substituting the values, A and B

C=[v(0)]2ω2+[x(0)]2

ϕ=90α

tanϕ=cotα=AB

tanϕ=v(0)x(0)ω

ϕ=tan1(v(0)x(0)ω)

Hence, (D) is the correct answer.

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