Question

A particle of mass $200g$ is making $SHM$ under the influence of a spring of force constant $k=90N/m$ and a damping constant $b=40g/s$ Calculate the time elapsed for the amplitude to drop to half its initial value [Given $ln(1/2)=0.693$ ]

• A. 7 s
• B. 9 s
• C. 4 s
• D. 11 s

Answer 1

Answer: (A)

The amplitude decreases continuously with time
$x=x_m{e}^{-(b/2m)t}$
$\frac{1}{2}{x}_m=x_m{e}^{-(40/2\times 200)t}$
$\frac{1}{2}=e^{-(t/10)}$
or $e^{(t/10)}=2$
$t/10=lo{g}_{e}2$
$t=10\times 0.693$
$t=7.00s$ (approximately)