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

x = A `cos(omegat + phi)" "…(i)`
x = 1 cm, V = `omegacms^(-1), t= pis^(-1)`
`phi = ?, A = ?`
Put value in (i)
`1 = A cos (pi xx 0 + phi) = A cos(phi)`
`A cos phi = 1 " "…(a)`
velocity ` = V = (dx)/(dt) = -Aomega sin(omegat + phi)`
At t = 0, V = `-Aomega sin phi`
At `V =omega, omega = -Aomegasinphi`
`rArr A sin phi =" "...(b)`
Squaring and adding (a) and (b) we get
`A^(2)(cos^(2)phi + sin^(2)phi) = 1 + 1`
`A = sqrt(2)cm`
Dividing we get
`(A sin phi)/(A cos phi) = -1`
`tan phi = -1`
`phi = (3pi)/(4) ~~ or (7pi)/(4)`
When `x = B sin(omegat + alpha) " "...(ii)`
At `t = 0, x = 1` we have
`1 = B sin alpha " "...(iii)`
Differntiating (ii) w.r.t. t we have
At t = 0, V = `omega, omega = pi`
`omega = Bomega cos (pi xx 0 + alpha)`
`1 = B cos alpha`
Squaring and adding (iii) and (iv)
`2 = B^(2) (sin^(2) alpha + cos^(2) alpha)`
`rArr B = sqrt(2)` cm
Dividing (iii) by (iv) we get
`(B sin alpha)/(B cos alpha) = (1)/(1) = tan alpha = 1, alpha = (pi)/(4)`
`alpha = (5pi)/(4)`