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(A) The equation of motion for a damped pendulum is given by $I \alpha = -mg \ell 	heta - b' v \ell$,where $b'$ is the damping constant. For a spheri

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$2/b$

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(A) The equation of motion for a damped pendulum is given by $I \alpha = -mg \ell 	heta - b' v \ell$,where $b'$ is the damping constant. For a spherical bob,the drag force is $F_d = -bv$. The angular displacement follows the form $	heta(t) = 	heta_0 e^{-(b/2m)t} \sin(\omega t + \phi)$.Given the amplitude decays as $A(t) = 	heta_0 e^{-(b/2m)t}$,we define the average life $	au$ as the time when the amplitude becomes $1/e$ of the initial amplitude $	heta_0$.Thus,$	heta_0/e = 	heta_0 e^{-(b/2m)	au}$.Comparing the exponents,we get $1 = (b/2m)	au$.Assuming the proportionality constant $b$ in the question refers to the damping factor per unit mass (often denoted as $b/m$),the average life is $	au = 2/b$.

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