Question

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 proportionality, the average life time of the pendulum in seconds (assuming damping is small) in second :-

• A. $\frac{0.693}{b}$
• B. $b$
• C. $\frac{1}{b}$
• D. $\frac{2}{b}$

Answer 1

Answer: (D)

$\frac{2}{b}$

Consider the free body diagram shown in the figure.

For the equilibrium in the normal direction, we have $T=mg\cos \theta$

In the tangential direction, $mL\ddot{\theta }=-mg\sin \theta -mbv$

For small angle $\theta$, we have $\sin \theta \approx \theta$

and $v=L\dot{\theta }$

Thus, $mL\ddot{\theta }+mbL\dot{\theta }+mg\theta =0$

Solution to the above differential equation is $\theta \left(t\right)={\theta }_0{e}^{-\frac{bt}{2}}\sin ({\omega }_{d}t+\varphi )$

Let average lifetime be $\tau$.

Amplitude at $t=0$ is ${\theta }_0$ and amplitude at$t=\tau$ is ${\theta }_0{e}^{-\frac{b\tau }{2}}$

But the ratio is equal to $1/e$

Thus, $e^{-\frac{b\tau }{2}}=e^{-1}\Rightarrow \tau =2/b$