A particle oscillates simple harmonic motion with a period of `16 s`. Two second after crossing the equilibrium position its velocity becomes `1 m//s`. The amplitude is

Options

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

Answer

B

Explanation

`omega = (2pi)/(T) = (2pi)/(16) = (pi)/(8) rad//s`
At `t = 0`, particle crosses the means position. Hence its velocity is maximum. So, velocity as a function of time can be written as
`v = v_(max) cos omega t`
or `v = omega A cos omega t`
`:. 1 = ((pi)/(8)) A cos ((pi)/(8))(2)`
`= ((pi)/(8sqrt(2)))A`
`:. A = (8sqrt(2))/(pi)m`