# 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 ? $(\ln 2=0.693)$

1. $34.65 \mathrm{~s}$

2. $17.32 \mathrm{~s}$

3. $0.034 \mathrm{~s}$

4. $15.01 \mathrm{~s}$

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