Q. The amplitude of wave disturbance propagating in positive direction of X –axis is given by $y = \frac{1}{1 + x^{2}}$ at t=0 and by $y = \frac{1}{1 + (x - 1)^{2}}$ at t=2s where $x\ $ and $y$ are in meters. The shape of the wave disturbance does not change during propagation. The velocity of the wave is
A) $0.5ms^{- 1}$
B) $2.0ms^{- 1}$
C) $1.0ms^{- 1}$
D) $4.0ms^{- 1}$
Explanation
(a) In a wave equation, $x$ and $t$ must be related in the form $(x - vt)$. Therefore, we rewrite the given equation as
$y = \frac{1}{1 + (x - vt)^{2}}$
For $t = 0,$ it becomes $y = \frac{1}{1 + x^{2}}$
And for $t = 2$, it becomes
$y = \frac{1}{\left\lbrack 1 + (x - 2v)^{2} \right\rbrack} = \frac{1}{1 + (x - 1)^{2}}$
$\therefore 2v = 1$ or $v = 0.5\ ms^{- 1}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.