The periodic time of a particle performing a linear S.H.M. is 12 sec. It starts from the mean position. After 2 seconds, its velocity is found to be `pi cm//sec`. What is the amplitude of S.H.M. ?

Options

A. 12 cm
B. 18 cm
C. 36 cm
D. 24 cm

Answer

D

Explanation

`omega =(2pi)/(T)=(2pi)/(12)=(pi)/(6)`
`x = A "sin" (omega t) therefore v= A omega "cos" omega t`
`therefore pi =A [(pi)/(6) "cos" ((pi)/(6)xx2)]`
`therefore 1 = (A)/(6) ("cos" 60^(@))=(A)/(12)`
`therefore A= 12 cm " and path length "= 2A=24 cm`