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. 1 cm
B. 2 cm
C. `pi cm`
D. `sqrt(2) cm`

Answer

D

Explanation

`x=A "cos" (omega t+ phi)`
When t=0, x=1 cm
`therefore 1= A "cos" phi `
and `v=(dx)/(dt)=(d)/(dt)(A "cos" omega t+ phi)`
`v= - A omega "sin"(omega t+ phi)`
`therefore " at " t=0, v= pi cm//s `(given)
`therefore pi = - A omega "sin" phi therefore (pi)/(omega)=- A "sin" phi`
But `omega =pi therefore (pi)/(pi)=1 = - A "sin" phi`
Squaring and adding (1) and (3), we get
`1+1=A^(2)("cos"^(2)phi+"sin"^(2)phi)=A^(2)`
`therefore A^(2)=2 or A=sqrt(2) cm`