If a simple pendulum has significant amplitude (up to a factor of $1 / e$ of original) only in the period between $t=0s$ to $t=\tau s$, then $\tau$ may be called the average life of the pendulum. When the spherical bob of the pendulum suffers a retardation (due to viscous drag) proportional to its velocity, with '$b$' as the constant of proportionality, the average life time of the pendulum is (assuming damping is small) in seconds:

• A

  $\frac{0.693}{b}$
• B

  $b$
• C

  $\frac{1}{b}$
• Correct answer
  D

  $\frac{2}{b}$

Solution

As retardation $=\mathrm{bv}$ $\therefore$ retarding force $=\mathrm{mbv}$ $\therefore$ net restoring torque when angular displacement is $\theta$ is given by $=-\operatorname{mg} \ell \sin \theta+\operatorname{mbv} \ell$ $\therefore \quad \mathrm{I} \alpha=-m g \ell \sin \theta+\operatorname{mbv} \ell$ where, $\mathrm{I}=\mathrm{m} \ell^2$ $\therefore \quad \frac{\mathrm{d}^2 \theta}{\mathrm{dt}^2}=\alpha=-\frac{\mathrm{g}}{\ell} \sin \theta+\frac{\mathrm{bv}}{\ell}$ for small damping, the solution of the above differential equation will be $\therefore \quad \theta=\theta_0 \mathrm{e}^{-\frac{\mathrm{bt}}{2}} \sin (w t+\phi)$ $\therefore \quad$ angular amplitude will be $=\theta . e^{\frac{-b t}{2}}$ According to question, in $\tau$ time (average life-time), angular amplitude drops to $\frac{1}{e}$ value of its original value $(\theta)$ $\therefore \quad \frac{\theta_0}{\mathrm{e}}=\theta_0 \mathrm{e}^{-\frac{6 \tau}{2}}$ $\frac{6 \tau}{2}=1$ $\therefore \quad \tau=\frac{2}{b}$

Hence, the correct answer is option: 'D'.