A block off mass 200 g executing SHM under the unfluence of a spring of spring constant `k=90Nm^(-1)` and a damping constant `gb=40gs^(-1)`. The time elaspsed for its amplitude to drop to halff of its initial value is (Given, ln `(1//2)=-0.693`)

Options

A. 7s
B. 9s
C. 4s
D. 11s

Answer

A

Explanation

The amplitude of the damped oscillator at any instant t is given by
`A(t)=A_(0)e^(-bt//2m)`
where `A_(0)` is its initial amplitude and b is the damping constant.
At `t=t_(1//2)`, the amplitude drop to half of its initial value.
From (i), we get
`(A_(0))/(2)=A_(0)e^(-bt_(1//2)//2m),(1)/(2)=e^(-bt_(1//2)/2m)`
Taking natural logarithm on both sides, we get
`ln((1)/(2))=-(bt_(1//2))/(2m),t_(1//2)=-(2mln(1//2))/(b)` . . . (ii)
here, `ln((1)/(2))=-0.693,b=40gs^(-1),m=200g`
substituting these values in eq. (ii) we get
`t_(1//2)=(0.693xx2xx200)/(40)s=7s`