A particle performing SHM is found at its equilibrium position at t = 1s and it is found to have a speed 0.25 = m//s at t=2s. If the period of oscillation is 6. Calculate amplitude of oscillation.

Options

A. `(3)/(2pi)m`
B. `(3)/(4pi)m`.
C. `(6)/(pi)m`
D. `(3)/(8pi)m`

Answer

A

Explanation

`x = a sin (omegat + phi)`
at `t = 1s, x = 0 = a sin(omegat + phi) rArr phi = -omega`
`:' v = a omega cos (omegat + phi)`
`:'` At `t = 2s, 1/4 = aomega cos (2omega + phi)`
`rArr 1/4 = a ((2pi)/(6)) cos (omega) = (api)/(3)cos ((pi)/(3)) = (api)/(3)((1)/(2)) rArr a = (3)/(2pi)`