The displacement of a particle in S.H.M. is given by `x=A "cos" (omegat+ phi)`
At time t=0, x=1 cm, initial velocity `=pi cm//s` and angular frequency is `pi` per second. What is the amplitude of SHM ?

Options

A. 2.5 cm
B. 2 cm
C. `sqrt(2) cm`
D. 1.5 cm

Answer

C

Explanation

`x=A "cos" (omega t+ phi)`
`therefore v=(dx)/(dt)= A omega "sin" (omega t+phi)`
at time `t=0, x=1 cm , v=pi cm//s and omega =pi rad//s`
U"sin"g these values in (1) and (2), we get
`1=A "cos" phi therefore "cos" phi =(1)/(A)`
`pi = - A xx pi "sin" (phi)`
`therefore "sin" phi = -(1)/(A) because "sin"^(2) phi + "cos"^(2) phi =1`
`therefore (1)/(A^(2))+(1)/(A^(2))=1 therefore (2)/(A^(2))=1`