A particle is performing a linear S.H.M. of period 24 s. The velocity of the particle is 6.28 m/s after 4 s after cros"sin"g the mean position. What is the amplitude of S.H.M. ?

Options

A. 48 cm
B. 36 cm
C. 24 cm
D. 12 cm

Answer

A

Explanation

`omega=(2pi)/(T)=(2pi)/(24)=(pi)/(12)`
`x = A "sin" omega t`
` therefore v= (dx)/(dt)= A omega "cos" omega t`
`therefore 6.28 = A xx (3.14)/(12) xx "cos" ((pi)/(12)xx4)`
`therefore 2xx12 =A "cos" (pi)/(3)=(A)/(2)`
`therefore A=48 cm`