The displacement of a damped oscillator is x( | vedclass

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(B) The displacement of a damped oscillator is given by $x(t) = A_0 e^{-bt} \cos(\omega' t + \Phi)$,where $A(t) = A_0 e^{-bt}$ is the amplitude at tim

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$6$

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(B) The displacement of a damped oscillator is given by $x(t) = A_0 e^{-bt} \cos(\omega' t + \Phi)$,where $A(t) = A_0 e^{-bt}$ is the amplitude at time $t$.Given the equation $x(t) = \exp(-0.2 t) \cos(3.2 t + \Phi)$,the initial amplitude $A_0$ at $t = 0$ is $1$.The amplitude at time $t$ is $A(t) = e^{-0.2 t}$.We need to find the time $t$ when the amplitude becomes $\frac{1}{e^{1.2}}$ times its initial amplitude.So,$A(t) = \frac{1}{e^{1.2}} 	imes A_0 = e^{-1.2} 	imes 1 = e^{-1.2}$.Equating the two expressions for amplitude: $e^{-0.2 t} = e^{-1.2}$.Comparing the exponents: $-0.2 t = -1.2$.Solving for $t$: $t = \frac{1.2}{0.2} = 6 \ s$.

The displacement of a damped oscillator is $x(t) = \exp(-0.2 t) \cos(3.2 t + \Phi)$,where $t$ is time in seconds. The time required for the amplitude of the oscillator to become $\frac{1}{e^{1.2}}$ times its initial amplitude is (in $s$)

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