If a simple pendulum has significant amplitude (up to a factor of 1 / e of original) only in the period between t=0s to t=τ s, then τ may be called…

As retardation $=\mathrm{bv}$=bv $\therefore$∴ retarding force $=\mathrm{mbv}$=mbv $\therefore$∴ net restoring torque when angular displacement is $\theta$θ is given by $=-\operatorname{mg} \ell \sin \theta+\operatorname{mbv} \ell$=−mgℓsinθ+mbvℓ $\therefore \quad \mathrm{I} \alpha=-m g \ell \sin \theta+\operatorname{mbv} \ell$∴Iα=−mgℓsinθ+mbvℓ where, $\mathrm{I}=\mathrm{m} \ell^2$I=mℓ2 $\therefore \quad \frac{\mathrm{d}^2 \theta}{\mathrm{dt}^2}=\alpha=-\frac{\mathrm{g}}{\ell} \sin \theta+\frac{\mathrm{bv}}{\ell}$∴dt2d2θ​=α=−ℓg​sinθ+ℓbv​ for small damping, the solution of the above differential equation will be $\therefore \quad \theta=\theta_0 \mathrm{e}^{-\frac{\mathrm{bt}}{2}} \sin (w t+\phi)$∴θ=θ0​e−2bt​sin(wt+ϕ) $\therefore \quad$∴ angular amplitude will be $=\theta . e^{\frac{-b t}{2}}$=θ.e2−bt​ According to question, in $\tau$τ time (average life-time), angular amplitude drops to $\frac{1}{e}$e1​ value of its original value $(\theta)$(θ) $\therefore \quad \frac{\theta_0}{\mathrm{e}}=\theta_0 \mathrm{e}^{-\frac{6 \tau}{2}}$∴eθ0​​=θ0​e−26τ​ $\frac{6 \tau}{2}=1$26τ​=1 $\therefore \quad \tau=\frac{2}{b}$∴τ=b2​