A body of mass 0.25 kg is attached to a vertical spring. The spring is executing damped simple harmonic spring will drop to half its initial value ? [The damping constant b = 0.05 kg/s]

Options

A. `4 "log"_(10^(2))`
B. `4 "log"_(e^(2))`
C. `5 "log"_(e^(2))`
D. `5 "log"_(10^(2))`

Answer

C

Explanation

m=0.25 kg and b =0.05 kg/s
The initial mechanical energy of a harmonic oscillator at time `t=0 is E_(1)=(1)/(2)KA^(2)`
But because of damping, its energy at time t becomes
`E_(2)=(1)/(2)KA^(2)e^((-bt)/(m))` where b is the damping constant.
It is given that at time `t, E_(2)=(E_(1))/(2)`
`therefore (E_(1))/(E_(2))=(1)/(e^(-bt)/(m))`
`therefore (E_(1))/((E_(1))/(2))=2=e^((bt)/(m))therefore(bt)/(m) = log_(e )2`
`therefore t= (m log_(e )2)/(b)=(0.25xx log_(e )2)/(0.05)`
`therefore t =5 log_(e )2`