The displacement of a particle in S.H.M. is given by `x= B "sin" (omega t + alpha)`. The initial position (at time t=0), of the particle is the initial phase angle if the angular frequency of the particle is `pi rad//s` ?

Options

A. `(pi)/(2)`
B. `(pi)/(4)`
C. `(pi)/(6)`
D. `(pi)/(5)`

Answer

B

Explanation

`x=B "sin" (omega t+ alpha)` At time t=0, x=1 cm
`therefore 1=B "sin" (omega xx 0+alpha)`
`therefore B "sin" alpha=1`
Similarly, `v=(dx)/(dt)= B omega "cos" (omega t+alpha)`
At time `t=0, v=pi cm//s`
` therefore pi=B (pi) "cos" (omega xx 0+alpha)`
`therefore B "cos" alpha =1`
`therefore ` From (i) and (ii),
`(B "sin" alpha)/(B "cos" alpha)=(1)/(1) therefore "tan" alpha=1 because "tan" 45^(@)=1`
`therefore alpha=45^(@) or (pi)/(4)`