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

For a damped harmonic oscillator,
Amplitude `(A)=A_(0)e^(-bt//2m)`
Suppose that the amplitude drops to half of its original value after time `t_(1)`.
` therefore (A_(0))/(2)=A_(0)e^((-bt_(1))/(2m)) therefore (1)/(2)=e^((-bt_(1))/(2m))`
`therefore` Taking natural logarithms on both sides, we get
`log_(e )((1)/(2))= - (bt_(1))/(2m)`
`therefore t_(1)=(-2m log_(e )((1)/(2)))/(b)`
But `log_(e )((1)/(2))= -0.693` (given)
m=200 g and b =40 g/s
`therefore` U"sin"g these values in (2), we get
` therefore t_(1)=(+2xx200xx0.693)/(40)=6.93=7s`