# Amplitude of a mass-spring system, which is

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 drop to half of its initial value ?

$(\ln 2=0.693)$

1. (1) $34.65 \mathrm{~s}$

2. (2) $17.32 \mathrm{~s}$

3. (3) $0.034 \mathrm{~s}$

4. (4) $15.01 \mathrm{~s}$

Correct Option: 1

Solution:

(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.