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)

$\frac{2}{b}$

Answer 2

$m \frac{d^{2}x}{dt^{2}}=-kx-b \frac{dx}{dt}$ $m \frac{d^{2}x}{dt^{2}}+b \frac{dx}{dt}+kx=0$ here $b$ is demping coefficient This has solution of type $x=e^{\lambda t}$ substituting this $m\lambda^{2}+b\lambda+k=0$ $\lambda=\frac{-b\pm\sqrt{b^{2}-4mk}}{2m}$ on solving for $x$, we get $x=e^{\frac{b}{2m}t} \,a cos \left(\omega_{1}\, t - \alpha\right)$ $\omega_{1}=\sqrt{\omega_{0}^{2}-\lambda^{2}}$ where $\omega_{0}=\sqrt{\frac{k}{m}}$ $\lambda=+\frac{b}{2}$ So, average life $=\frac{2}{b}$