The time period and amplitude of a particle executing SHM are 10 s and 0.12 m respectively. Find its velocity at a distance 0.04 m from its position of equilibrium.

Explanation

Here T = 10 s : A = 0.12 m
`therefore" "omega=(2pi)/T=(2pi)/10=pi/5"rad"*s^(-1)`
So, `v=omegasqrt(A^(2)-x^(2))=pi/5sqrt(0.12^(2)-0.04^(2))=pi/5sqrt(0.0128)`
`=0.071m*s^(-1)`.