The motion of a particle executing simple harmonic motion is described by the displacement function, . If the initial position of the particle is 1 cm and its initial velocity is , what are its amplitude and initial phase angle?
The angular frequency of the particle is . If instead of the cosine function, we choose the sine function to describe the SHM: , what are the amplitude and initial phase of the particle with the above initial conditions?

Explanation

Here , at t=0 , x=1 cm and `V = omega cm s^(-1) , omega = pi s^(-1)` Given x=A `cos (omega t +phi)`
`therefore 1 = A cos (pi xx 0 + phi) = A cos phi" "….(i) ` Velocity , `V = (dx)/(dt) =- A omega sin (omega t + phi)`
`therefore =- A omega sin (pi xx 0 + phi) ` (or) `1=-Asin phi ` (or) `A sin phi =-1 " "....(ii)`
Squaring and adding (i) and (ii) `A^2 cos^2 phi + sin^2 phi) = 1+2` (or)` A^2 =2 ` (or) `A = sqrt2 ` cm
Dividing (ii) by (i) we get `tan phi` =-1 (or) `phi = (3pi)/(4) ` (or)` (7pi)/(4) ` For , `x= B sin (omega t + alpha)" ".....(iii)`
At t=0 , x=1 , so, 1= Bsin `(omega xx 0 + alpha)= B sin alpha " "....(iv)`
Differentiating (iii), w.r.t.t, we have velocity , `V= (dx)/(dt) = B omega cos (omega t+ alpha)`
Applying initial conditions i.e., at `t=0 , V= omega`
`omega = B omega cos (pi xx 0 + alpha)` or `1= B cos alpha" "....(iv)`
Squaring and adding (iv) and(v) , we get `B^2 sin^2 alpha + B^2 cos^2 alpha = 1^2 + 1^2 = 2 `or `B^2 = 2 ` or `B= sqrt2 ` cm
Dividing (iv) by (v) , we have `(B sin alpha)/(B cos alpha) = 1/1 ` (or) `tan alpha = 1 `(or)` alpha = pi//4 ` or `5pi//4`.