Oscillations

JEE Main level, single correct, medium

165 of 194

Question 165 of 194

The amplitude of a mass-spring system, which is executing simple harmonic motion decreases with time. If mass =500g=500g, Decay constant =20gs−1=20g{s}^{-1} then how much time is required for the amplitude of the system to drop to half of its initial value? (ln⁡2=0.693)\left(\ln 2=0.693\right)

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Answer: (A) 34.65s34.65s

A=A0e−γt=A0e−bt2mA={A}_{0}{e}^{-\gamma t}={A}_{0}{e}^{-\dfrac{bt}{2m}}
A02=A0e−bt2m\dfrac{{A}_{0}}{2}={A}_{0}{e}^{-\dfrac{bt}{2m}}
bt2m=ln⁡2\dfrac{bt}{2m}=\ln 2
t=2mbln⁡2=2×500×0.69320t=\dfrac{2m}{b}\ln 2=\dfrac{2\times 500\times 0.693}{20}
t=34.65t=34.65 second.

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