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(A) For a damped harmonic oscillator,the amplitude $A$ at time $t$ is given by $A = A_0 e^{-\frac{bt}{2m}}$.We want to find the time $t$ when the ampl

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$34.65$

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(A) For a damped harmonic oscillator,the amplitude $A$ at time $t$ is given by $A = A_0 e^{-\frac{bt}{2m}}$.We want to find the time $t$ when the amplitude drops to half of its initial value,i.e.,$A = \frac{A_0}{2}$.Substituting this into the equation: $\frac{A_0}{2} = A_0 e^{-\frac{bt}{2m}}$.This simplifies to $\frac{1}{2} = e^{-\frac{bt}{2m}}$,or $2 = e^{\frac{bt}{2m}}$.Taking the natural logarithm on both sides: $\ln 2 = \frac{bt}{2m}$.Solving for $t$: $t = \frac{2m}{b} \ln 2$.Given $m = 500 \, g$,$b = 20 \, g/s$,and $\ln 2 = 0.693$:$t = \frac{2 	imes 500}{20} 	imes 0.693 = 50 	imes 0.693 = 34.65 \, s$.

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