For a damped oscillation of a particle, show that time taken for the amplitude to drop to half of its initial value `=(2m ln (2))/(b)`, where b is a damping constant.

Explanation

Instantaneous displacement of a particle `=x(t)=Ae^(-bt//2m)cos(omega^(1)t+phi)`
Taking `cos(omega.t+phi)=1" for "omega.t+pi=0 and x(t)=(A)/(2)`
We get`" "(A)/(2)=(A)/(e^(bt//2m))i.e. e^(bt//2m)=2`
`therefore" "t=(2.303log2)(2m)/(b)=(2m ln 2)/(b)`