A bob executes shm of period 20 s. Its velocity is found to be `0.05 ms^(-1)` after 2 s when it has passed through its mean position. Find the amplitude of the bob.

Explanation

`T = 20 s, v=0.05 ms^(-1), t=2 s`
`omega = (2pi)/T =2/20 pi = (pi)/10 rad//s`
`v = aomega cos omegat`
`=0.05 = (api)/10 cos""(pi)/10 xx2 = (pia)/10 cos""pi/5`
`0.05 = (pia)/10 xx0.809`
`a= (0.05xx10)/(3.14 xx0.809)=0.197m`