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

The given displacement function is
`x (t) = A cos (omega t +phi)` ...(i)
At `t = 0, x(0) = 1 cm`, Also `omega = pi s^(-1)`
`:. 1 = A cos (pi xx 0 +phi)`
`rArr A cos phi = 1` ...(ii)
Also, differentiating eqn (i) w.r.t 't'
`v = (d)/(dt) x (t) =- A omega sin (omega t +phi)` ....(iii)
Now at `t = 0, v = omega`
`:.` from eq. (iii), `omega = - A omega sin (pi xx 0 + phi)`
or `A sin phi =- 1`
Squaring and adding eqns. (ii) and (iv) ...(iv)
`A^(2) cos^(2) phi + A^(2) sin^(2) phi = 1^(2) + 1^(2)` or `A = sqrt(2) cm`
Dividing eqns. (ii) and (iv),
`(A sin phi)/(A cos phi) = (-1)/(1) :. tan phi =- 1 rArr phi = (3pi)/(4)`
If instead we use the sine function, i.e.,
`x = B sin (omega t +alpha)`, then `v = (d)/(dt) B omega cos (omega t +alpha)`
`:.` At `t = 0`, using `x = 1` and `v = omega`, we get `1 = B sin (omega xx 0 + alpha)`
or `B sin alpha = 1` ...(v)
and `omega = B omega (omega xx 0 + alpha)` or `B cos alpha = 1` ...(vi)
Dividing (v) by (vi),
`tan alpha = 1` or `alpha = (pi)/(4)` or `(5pi)/(4)`
`B^(2) sin^(2) alpha + B^(2) cos^(2) alpha = 1^(2) +1^(2)`
`rArr B = sqrt(2) cm`.