Amplitude of a mass-spring system, which is
executing simple harmonic motion decreases with time. If
mass $=500 \mathrm{~g}$, Decay constant $=20 \mathrm{~g} / \mathrm{s}$ then how much time is required for the amplitude of the system to drop to half of its initial value ?
$(\ln 2=0.693)$
Correct Option: 1
(1)
$\mathrm{A}=\mathrm{A}_{0} \mathrm{e}^{-\gamma \mathrm{t}}=\mathrm{A}_{0} \mathrm{e}^{-\frac{\mathrm{bt}}{2 \mathrm{~m}}}$
$\frac{\mathrm{A}_{0}}{2}=\mathrm{A}_{0} \mathrm{e}^{-\frac{\mathrm{bt}}{2 \mathrm{~m}}}$
$\frac{\mathrm{bt}}{2 \mathrm{~m}}=\ln 2$
$\mathrm{t}=\frac{2 \mathrm{~m}}{\mathrm{~b}} \ln 2=\frac{2 \times 500 \times 0.693}{20}$
$\mathrm{t}=34.65$ second.
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