The motion of a particle executing simple harmonic motion is described by the displacement function,
`x(t)=Acos(omegat+phi)`.
If the initial (t = 0) position of the particle is 1 cm and its initial velocity is `omega` cm/s, what are its amplitude and initial phase angle ? The angular frequency of the particle is `pi s^(-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(t) =A cos (omegat( +phi)`
`x(0) = A cos phi" "[therefore t= 0]`
`therefore 1= A cos phi " ""………."(1)" "[x(0)= 1 cm]`
Also, diffrerentiating this equation. `omega.r.t. .t.`.
`v(t) = - A omega sin(omega t+phi)`
`therefore v(0) = -A omega sin phi " "[therefore t=0]`
`therefore omega = -A omega sin phi " "[therefore v(0) = omega]`
`therefore 1= -A sin phi" ""........"(2)`
Squaring and adding equn. (1) and (2)
`1+1 = A^(2) cos^(2) phi +A^(2) sin^(2) phi`
`therefore 2 = A^(2)" "[therefore cos^(2) phi+ sin^(2) phi =1]`
`therefore A = pm sqrt(2) cm`
Dividing eqn. (2) by (1),
`-1 = tan phi implies tan phi = -1`
`therefore tan phi = -tan (pi)/(4)`
`tan phi = tan(2pi -(pi)/(4))`
`therefore phi = 2pi -(pi)/(4)`
`therefore phi = (7pi)/(4)rad`
Now let consider `x(t) = B sin (omega t + alpha)`
`x(0) = B sin alpha " " [therefore t=0]`
`1= B sin alpha " ""........"(3)[therefore x(0) =1]`
and differentiating this equation `omega.r.t. .t.`,
`v(t)= B omega cos (omega t+alpha)`
`v(0) = B omega cos alpha" "[therefore t=0]`
`therefore omega = B omega cos alpha " "[therefore v(0) = omega]`
`therefore 1= B cos alpha " ""......"(4)`
Squaring and adding equn. (3) and (4),
`1^(2)+1^(2) = B^(2)sin^(2) alpha+ B^(2) cos^(2) alpha`
`therefore 1+1 = B^(2)[ sin^(2) alpha+ cos^(2) alpha]`
`therefore 2 = B^(2)" "[therefore sin^(2)alpha+ cos^(2)alpha= 1]`
`therefore B = pm sqrt(2) cm`
Diving (3) and (4),
`1 = tan alpha`
`therefore alpha = tan^(-1) (1)`
`therefore alpha = (pi)/(4)rad`