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 ?

Explanation

Given at t = 0, y = 1 c.m.
`therefore" "1 c.m = A cos (omega xx 0 + phi)`
`rArr" "1 = A cos phi" "...(1)`
At t = 0, `v = omega` cm/s
velocity of the particle
`v = - A omega sin (omega t + phi)`
`rArr" "omega =-A omega sin (omega xx 0 + phi)`
`rArr" "-1 = A sin (phi)" "...(ii)`
From equation (i) and (ii)
`(A sin varphi)/(A cos varphi) =-1`
`rArr tan phi =-1 rArr phi = (3pi)/(4)`
Substituting the value of `phi` in equation (i), we get
`1 = A "cos" (3pi)/(4)`
`A = sqrt(2) cm`