The amplitude of a simple pendulum,oscillatin | vedclass

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(C) The amplitude of a damped oscillator is given by $A(t) = A_0 e^{-(b/2m)t}$,where $b$ is the damping constant. According to Stokes' law,$b = 6\pi \

Vedclass · Accepted Answer

$161$

Vedclass · Answer

(C) The amplitude of a damped oscillator is given by $A(t) = A_0 e^{-(b/2m)t}$,where $b$ is the damping constant. According to Stokes' law,$b = 6\pi \eta r$,where $\eta$ is the coefficient of viscosity.For air: $8 = 10 e^{-(b_{air}/2m) \cdot 40} \implies 0.8 = e^{-(b_{air}/2m) \cdot 40}$.Taking natural log: $\ln(0.8) = -(b_{air}/2m) \cdot 40 \implies \ln(4/5) = -(b_{air}/2m) \cdot 40 \implies \ln(5/4) = (b_{air}/2m) \cdot 40$.So,$(b_{air}/2m) = \frac{\ln(1.25)}{40} = \frac{0.223}{40} = 0.005575 \ s^{-1}$.Given $\frac{\eta_{air}}{\eta_{CO_2}} = 1.3$,then $b_{CO_2} = \frac{b_{air}}{1.3}$.Thus,$(b_{CO_2}/2m) = \frac{b_{air}}{1.3 \cdot 2m} = \frac{0.005575}{1.3} \approx 0.004288 \ s^{-1}$.For $CO_2$: $5 = 10 e^{-(b_{CO_2}/2m) \cdot t} \implies 0.5 = e^{-(0.004288)t}$.Taking natural log: $\ln(0.5) = -0.004288 \cdot t \implies -0.693 = -0.004288 \cdot t$.$t = \frac{0.693}{0.004288} \approx 161.6 \ s$.Therefore,the time is close to $161 \ s$.

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