Wave Mechanics Question 607
Question: The amplitude of a damped oscillator becomes $ {{( \frac{1}{3} )}^{rd}} $ in 2 seconds. If its amplitude after 6 seconds is $ \frac{1}{n} $ times the original amplitude, the value of n is
Options:
A) $ 3^{2} $
B) $ 3^{3} $
C) $ \sqrt[3]{3} $
D) $ 2^{3} $
Show Answer
Answer:
Correct Answer: B
Solution:
[b] Amplitude of a damped oscillator at any instant t is given by $ A=A _{0}{{e}^{-bt/2m}} $ where $ A _{0} $ is the original amplitude From question, When $ t=2s,,A=\frac{A _{0}}{3}\
Therefore ,\frac{A _{0}}{3}=A _{0}{{e}^{-2b/2m}} $ or $ \frac{1}{3}={{e}^{-b/m}} $ ?. (i) When $ t=6s,,A=\frac{A _{0}}{n} $
$ \
Therefore ,\frac{A _{0}}{n}=A _{0}{{e}^{-6b/2m}} $ or, $ \frac{1}{n}={{e}^{-3b/m}}={{({{e}^{-b/m}})}^{3}} $ or, $ \frac{1}{n}={{( \frac{1}{3} )}^{3}} $
$ \
Therefore ,n=3^{3} $ (Using eq. (i))
BETA
