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

Amplitude of damped oscillator-
`A(t)=A_(0)e^(-bt//2m)`
`implies (A_(0))/(2)=A_(0)e^((-bt)/(2m))implies ln((1)/(2))=(-bt)/(2m)`
`implies -0.693=(-40xx10^(-3)xxt)/(2xx0.2)`
`implies t=(2xx0.2xx0.693)/(40xx10^(-3))=6.93` sec
`implies t~~7` sec.