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

Answer

D

Explanation

Given , `v = pi m//s, T = 16s`
Displacement of the particle `= v = (dx)/(dt) = A omega cos omegat"…"(i)`
Angular velocity, `omega = (2pi)/(T) = (2pi)/(16) = (pi)/(8)` rad/s
Substituting the values in Eq. we get
`pi = A xx pi/8 xx cos ' pi/8 xx 2`
gt `rArr 1 = A/8 cos'pi/4`
`= A/(8).(1)/(sqrt(2))`
` A = 8sqrt(2) m`