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 of proportionality, the averatge life time of the pendulum is (assuming damping is small)

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

Answer 1

Answer: (D)

Consider the free body diagram shown in the figure.
For the equilibrium in the normal direction, we have $T=m g \cos \theta$
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 $\theta(t)=\theta_0 e^{-\frac{b t}{2}} \sin \left(\omega_{ d } t +\phi\right)$
Let average lifetime be $\tau$.
Amplitude at $t =0$ is $\theta_0$ and amplitude at $=\tau$ is $\theta_0 e ^{-\frac{ b \tau}{2}}$
But the ratio is equal to $1 / e$
Thus, $e ^{-\frac{ b \tau}{2}}= e ^{-1} \Rightarrow \tau=2 / b$