A particle performing SHM starts equilibrium position and its time period is 16 seconds. After 2 seconds its velocity is `pi m//s`. Amplitude of oscillation is
`(cos 45^(@) = 1/(sqrt(2)))`

Options

A. `2 sqrt(2)m`
B. `4sqrt(2) m`
C. `6sqrt(2) m`
D. `8sqrt(2)m`

Answer

D

Explanation

The particle starts from rest. The displacement of the S.H.M. particle `=x=A "sin" omega t`
and its velocity `=v=(dx)/(dt)=A omega "cos" omega t`
In this case, `omega =(2pi)/(T)=(2pi)/(16)=(pi)/(8) rad//s and v=pi m//s`
U"sin"g (1), we get
`pi=A ((pi)/(8)) "cos" ((pi)/(8))xx2`
`therefore 1=(A)/(8) "cos" (pi)/(4)=(A)/(8)xx(1)/(sqrt(2))=(A)/(8sqrt(2))`
`therefore A=8sqrt(2)m`