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lfa simple pendulum has significant amplitude (up to a factor of $1/e$ of original) only in the period between $t = 0\ s$ 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
$\frac{2}{b}$
$\frac{{0.693}}{b}$
$b$
$\frac{1}{b}$.
Solution
The equation of motion for the pendulum, suffering retardation
$I \alpha=-m g(\ell \sin \theta)-m b v(\ell)$ where $I=m \ell^{2}$
and $\alpha=\mathrm{d}^{2} \theta / d t^{2}$
$\therefore \frac{d^{2} \theta}{d t^{2}}=-\frac{g}{\ell} \tan \theta+\frac{b v}{\ell}$
On solving we get $\theta=\theta_{0} e^{-\frac{b t}{2} \sin (\omega t+\phi)}$
According to questions $\frac{\theta_{0}}{e}=\theta_{0} e^{\frac{-b \tau}{2}}$
$\therefore \tau=\frac{2}{b}$
