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(B) The amplitude of a damped oscillator is given by $A(t) = A_0 e^{-bt / 2m}$.Mechanical energy $E$ is proportional to the square of the amplitude, $

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$4.0$

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(B) The amplitude of a damped oscillator is given by $A(t) = A_0 e^{-bt / 2m}$.Mechanical energy $E$ is proportional to the square of the amplitude, $E \propto A^2$.Therefore, $E(t) = E_0 e^{-bt / m}$.We want to find the time $t$ when $E(t) = E_0 / 4$.Substituting this into the equation: $E_0 / 4 = E_0 e^{-bt / m}$.This simplifies to $1/4 = e^{-bt / m}$, or $2^{-2} = e^{-bt / m}$.Taking the natural logarithm on both sides: $-2 \ln 2 = -bt / m$.Solving for $t$: $t = (2m \ln 2) / b$.Given $m = 200 	ext{ g} = 0.2 	ext{ kg}$, $b = 70 	ext{ g/s} = 0.07 	ext{ kg/s}$, and $\ln 2 = 0.7$.$t = (2 	imes 0.2 	imes 0.7) / 0.07 = 0.28 / 0.07 = 4 	ext{ s}$.

The amplitude of a damped oscillator varies with time as $A(t) = A_0 \exp(-bt / 2m)$, where $b = 70 \text{ g/s}$ and $m = 200 \text{ g}$. How long does it take for the mechanical energy to drop to one-fourth of its initial value (in $s$)? (Take $\ln 2 = 0.7$)

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