Wave Mechanics Question 604
Question: If a simple pendulum has significant amplitude (upto 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 average life time of the pendulum is (assuming damping the small) in seconds
Options:
A) $ \frac{0.693}{b} $
B) b
C) $ \frac{1}{b} $
D) $ \frac{2}{b} $
Show Answer
Answer:
Correct Answer: D
Solution:
[d] For damped harmonic motion, $ ma=-kx-mbv $ or $ ma+mbv+kx=0 $ Solution to above equation is $ x=A _{0}{{e}^{-\frac{bt}{2}}},\sin ,\omega t; $ with $ {{\omega }^{2}}=\frac{k}{m}-\frac{b^{2}}{4} $ where amplitude drops exponentially with time i.e., $ {A _{\tau }}=A _{0}{{e}^{-\frac{b\tau }{2}}} $ Average time $ \tau $ is that duration when amplitude drops by 63%, i.e., becomes $ A _{0}/e $ . Thus, $ {A _{\tau }}=\frac{A _{0}}{e}=A _{0}{{e}^{-\frac{b\tau }{2}}} $ or $ \frac{b\tau }{2}=1,or,\tau =\frac{2}{b} $
BETA
