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The displacement of a damped harmonic oscillator is x(t) = e^(-0.1*t) * cos(10*pi*t + phi), with t in seconds. Approximately how long does it take for the amplitude of vibration to fall to half its initial value?

1. 13 s
2. 7 s
3. 27 s
4. 4 s

Correct answer: 7 s

Solution

The amplitude decays as A(t) = A0*e^(-0.1*t). Setting A(t) = A0/2 gives e^(-0.1*t) = 1/2, so 0.1*t = ln2 = 0.693, hence t = 0.693/0.1 = 6.93 s, approximately 7 s.

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