Question

if a simple pendulum has significant amplitude( up to a factor of 1/e of original) only in a peroid between t=0s to t=Tsec then T 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 into velocity with b as the constant of proportionality the average life time of the pendulum is (assuming damping is small) in seconds

Answer 1

Dear Student,

$$\begin{gathered} \mathrm{As}\mathrm{retardation}=\mathrm{bv} \\ \mathrm{retarding}\mathrm{force}=\mathrm{mbv} \\ \mathrm{net}\mathrm{restoring}\mathrm{torque}\mathrm{when}\mathrm{angular}\mathrm{displacement}\mathrm{is}\mathrm{\theta }\mathrm{is}, \\ \mathrm{\tau }=-\mathrm{mgl}\mathrm{sin\theta }+\mathrm{mbvl} \\ \mathrm{I\alpha }=-\mathrm{mgl}\mathrm{sin\theta }+\mathrm{mbvl}........\left[\mathrm{because}\mathrm{\tau }=\mathrm{I\alpha }\right]........\left(1\right) \\ \mathrm{Also},\mathrm{I}={\mathrm{ml}}^2...\left(2\right) \\ \mathrm{from}\left(1\right)\&\left(2\right), \\ \frac{{\mathrm{d}}^2\mathrm{\theta }}{{\mathrm{dt}}^2}=\mathrm{\alpha }=-\frac{\mathrm{g}}{\mathrm{l}}\mathrm{sin\theta }+\frac{\mathrm{bv}}{\mathrm{l}} \\ \mathrm{for}\mathrm{small}\mathrm{damping},\mathrm{the}\mathrm{solution}\mathrm{of}\mathrm{the}\mathrm{above}\mathrm{differential}\mathrm{equation}\mathrm{will}\mathrm{be} \\ \mathrm{\theta }={\mathrm{\theta }}_0{\mathrm{e}}^{\frac{\mathrm{bt}}{2}}\sin \left(\mathrm{\omega t}+\mathrm{\varphi }\right) \\ \mathrm{therefore},\mathrm{angular}\mathrm{amplitude}\mathrm{will}\mathrm{be}={\mathrm{\theta e}}^{\frac{-\mathrm{bt}}{2}} \\ \mathrm{According}\mathrm{to}\mathrm{question},\mathrm{in}\mathrm{t}\mathrm{time}\left(\mathrm{average}\mathrm{life}-\mathrm{time}\right), \\ \mathrm{angular}\mathrm{amplitude}\mathrm{drops}\mathrm{to}\frac{1}{\mathrm{e}}\mathrm{value}\mathrm{of}\mathrm{its}\mathrm{original}\mathrm{value}\left(\mathrm{\theta }\right) \\ \frac{{\mathrm{\theta }}_0}{\mathrm{e}}={\mathrm{\theta }}_0{\mathrm{e}}^{\frac{\mathrm{bt}}{2}} \\ \frac{\mathrm{bt}}{2}=1 \\ \mathrm{t}=\frac{2}{\mathrm{b}} \end{gathered}$$