Question
If a simple pendulum has significant amplitude (upto a factor of $$\frac{1}{e}$$ 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 the small) in seconds
A.
$$\frac{{0.693}}{b}$$
B.
$$b$$
C.
$$\frac{1}{b}$$
D.
$$\frac{2}{b}$$
Answer :
$$\frac{2}{b}$$
Solution :
For damped harmonic motion,
$$ma = - kx - mbv\,\,{\text{or}}\,\,ma + mbv + kx = 0$$
Solution to above equation is
$$x = {A_0}{e^{ - \frac{{bt}}{2}}}\sin \,\omega t;\,\,{\text{with}}\,\,{\omega ^2} - \frac{k}{m} - \frac{{{b^2}}}{4}$$
where amplitude drops exponentially with time
$${\text{i}}{\text{.e}}{\text{.,}}\,\,{A_\tau } = {A_0}{e^{ - \frac{{b\tau }}{2}}}$$
Average time $$\tau $$ is that duration when amplitude drops by $$63\% ,$$ i.e., becomes $$\frac{{{A_0}}}{e}.$$
$$\eqalign{
& {\text{Thus,}}\,\,{A_\tau } = \frac{{{A_0}}}{e} = {A_0}{e^{ - \frac{{b\tau }}{2}}} \cr
& {\text{or}}\,\,\frac{{b\tau }}{2} = 1\,\,{\text{or}}\,\,\tau = \frac{2}{b} \cr} $$