A particle is `SHM` is discribed by the displacement function `x(t) = a cos (Delta omega + theta)` If the initial `(t = 0)` position of the particle `1cm` and its initial velocity is `picm//s` The angular frequency of the particle is `pi rad//s`, then its amplitude is

Options

A. `1cm`
B. `sqrt(2)cm`
C. `2 cm`
D. `2.5 cm`

Answer

B

Explanation

`x = a cos(omega t +theta)`…(i)
And `v = (dx)/(dt) = - aomegasin(omega t + theta)`…(ii)
Given at `t = 0 , x = 1 cm and v = pi` and `omega = pi`
Putting these value in equation (i) and (ii) , are will get
`sin theta = (-1)/(a)` and `cos theta = (1)/(A)`
`rArr sin^(2) theta + cos^(2) theta = (-(1)/(a))^(2) +((1)/(a))^(2) rArr a = sqrt(2) cm`