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The displacement of a damped harmonic oscillator is given by $x (t) = e^{- 0.1 t} \cos (10 π t + φ)$ , here t is in seconds. The time taken in seconds for its amplitude of vibration to drop to 1/4th of its initial value is close to____( $\ln 2 = 0.693$ )

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answer is 14.

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Detailed Solution

Amplitude of vibration at time t=0 is given by

$A = A_{0} e^{- 0.1 \times 0} = 1 \times A_{0} = A_{0} h e r e A_{0} = 1$
Also at $t = t$ ,

if $A = \frac{A_{0}}{4} \Rightarrow \frac{1}{4} = e^{- 0.1 t} \Rightarrow t = 10 I n 2^{2} = 14 \sec$

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