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

AMUAMU 2012Oscillations

Solution:

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 $e^{(t/10)} = 2$
$t/10 = log_e \,2$
$t = 10\times 0.693$
$ t = 7.00\,s$ (approximately)