Wave Mechanics Question 626
Question: The amplitude of a wave disturbance propagating in the positive x-direction is given by $ y=\frac{1}{1+x^{2}} $ at t=0 and $ y=\frac{1}{2+x^{2}-2x} $ at t = 2s, where x and y are in meter. Assuming that the shape of the wave disturbance does not change during the propagation, the speed of the wave is
Options:
A) $ 0.5m/s $
B) $ ~1m/s $
C) $ 1.5m/s $
D) $ 2m/s $
Show Answer
Answer:
Correct Answer: A
Solution:
[a] At, $ t=0,,y=\frac{1}{1+x^{2}}, $ or $ x=\sqrt{\frac{1-y}{y}}=x _{1} $ At $ t=2s, $
$ y=\frac{1}{2+x^{2}-2x}=\frac{1}{1+{{(x-1)}^{2}}} $ or $ {{(x-1)}^{2}}=\frac{1-y}{y} $ or $ x=1+\sqrt{\frac{1-y}{y}}=x _{2} $
$ \
Therefore Speedofthewave $
$ v=\frac{\Delta x}{\Delta t}=\frac{x _{2}-x _{1}}{t _{2}-t _{1}}=\frac{1}{2-0}=0.5m/s $
BETA
