Amplitude of a mass-spring system, which is executing simple harmonic motion decreases with time.

Question:
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 dropAmplitude 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 ? $(\ln 2=0.693)$

$34.65 \mathrm{~s}$
$17.32 \mathrm{~s}$
$0.034 \mathrm{~s}$
$15.01 \mathrm{~s}$

Correct Option:

Solution:

$\mathrm{A}=\mathrm{A}_{0} \mathrm{e}^{-\gamma \mathrm{t}}=\mathrm{A}_{0} \mathrm{e}^{-\frac{\mathrm{bt}}{2 \mathrm{~m}}}$

$\frac{A_{0}}{2}=A_{0} e^{-\frac{b t}{2 m}}$

$\frac{b t}{2 m}=\ln 2$

$t=\frac{2 m}{b} \ln 2=\frac{2 \times 500 \times 0.693}{20}$

$\mathrm{t}=34.65$ second

Amplitude of a mass-spring system, which is executing simple harmonic motion decreases with time.

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