> Q. For a damped harmonic oscillator of mass $200, g$, the values of spring constant and damping constant are, respectively, $90, N/m$ and $0.04, kg/s$. The time taken for its amplitude of vibration of drop to half of its initial value is (log, 2 = 0.693) – LIVE ANSWER TODAY

Q. For a damped harmonic oscillator of mass $200, g$, the values of spring constant and damping constant are, respectively, $90, N/m$ and $0.04, kg/s$. The time taken for its amplitude of vibration of drop to half of its initial value is (log, 2 = 0.693)

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For a damped harmonic oscillator of mass $200, g$, the values of spring constant and damping constant are, respectively, $90, N/m$ and $0.04, kg/s$. The time taken for its amplitude of vibration of drop to half of its initial value is (log, 2 = 0.693)

Solution:

The given, $ K=90,N/m $ $ b=0.04,kg/s $ $ {{log }_{e}}=2=0.693 $ $ m=200,g=0.2,kg $ The (instantaneous) amplitude of damped oscillations is given by $ {{A}_{t}}=A{{e}^{-bt/2m}} $ Let $ {{t}_{1}} $ be the time taken for the amplitude to drop to half of its initial value, that is at $ t={{t}_{1}} $ , $ {{A}_{t}}=A/2 $ Thus, $ frac{A}{2}=A{{e}^{-b{{t}_{1}}/2m}} $ $ {{e}^{b{{t}_{1}}/2m}}=2 $ $ frac{b{{t}_{1}}}{2m}={{log }_{e}}2 $ $ {{t}_{1}}=frac{2m}{b}{{log }_{e}}2 $ $ =frac{2times 0.2}{0.04}{{log }_{e}}2=frac{0.4}{0.04}times 0.693 $ $ =10times 0.693=6.93,s $ or $ cong $ 7.0 s

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