A particle is SHM is described by the displacement function.
`x = A cos(omega t + phi), "where," omega = 2pi//T`
If the initial (t = 0) position of the particle is 1 cm and its initial velocity is `pi cms^(-1)`, what is the initial phase angle ? The angular frequency of the particle is `pi s^(-1)`.

Options

A. `3 pi//4`
B. `2 pi//4`
C. `5 pi//4`
D. `7 pi//4`

Answer

A

Explanation

Here, at t = 0, x = 1 cm and `v = pi cms^(-1), phi = ?, omega = pi s^(-1)`
Given, `x = A cos (omegat + phi)`
`1 = A cos (pi xx 0 + phi)=A cos phi" "...(i)`
Velocity, `v=(dx)/(dt)=-A omega sin(omega t+phi)`
`therefore" "pi=-A pi sin(pi xx 0 + phi)=-A pi sin phi" "...(ii)`
or `1=-A sin phi rArr A sin phi =-1`
Dividing Eq. (ii) by Eq. (i), we get
`tan phi =-1 =-"tan"(pi)/(4)=tan(pi-(pi)/(4))="tan"(3pi)/(4)or phi=(3pi)/(4)`