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The motion of a particle in S.H.M. is de...

The motion of a particle in S.H.M. is described by the displacement function, `x=Acos(omegat+phi)`, If the initial `(t=0)` position of the particle is 1cm and its initial velocity is `omega cm s^(-1)`, what are its amplitude and initial phase angle ? The angular frequency of the particle is `pis^(-1)`. If instead of the cosine function, we choose the sine function to describe the SHM`: x=B sin(omegat+alpha)`, what are the amplitude and initial phase of the particle with the above initial conditions ?

Text Solution

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Initially, at t=0:
Displacement , x=1 cm
Initial velocity, `v=omegacm//sec`
Anugular frequency, `omega=pirad//s^(-1)`
It is given that : `x(t)=Acos(omegat+phi)`
`1=Acos(omega xx o+phi)=Acosphi`
`Acosphi=1`…......(i)
Velocity,` v=(dx)/(dt)`
`omega=Aomegasin(omegat+phi)`
`1=-A(omega xx 0 +phi)=-Asinphi`
A `
sinphi =-1........(ii)
`Squaring and adding equaations (i)and (ii), we get :
`A^(2)(sin^(2)phi+cos^(2)phi)=+1+1`
`A^(2)=2`
`A=sqrt2 cm`
Dividing equation (ii)by equation (i),we get:
`tanphi=-1`
`therefore phi =(3pi)/(4)`,`(7pi)/(4)`........
SMH is given as:`x=BomegaB cos(omegat+a)`
Substituting the given values, we get:
`pi=piBsina`
Bsina=1........(iv)
Squaring nd adding equations(iii)and (iv),we get:
`B^(2)[sin^(2)a+cos^(2)a]=1+1`
`B^(2)=2`
`B=sqrt2cm`
Dividing equation (iii)by equation (iv),weget:
`(Bsina)/(Bcosa)=(1)/(1)`, `tan a=1= "tan"(pi)/(4)` `thereforea=(pi)/(4)`,`(5pi)/(4)`.........
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