The displacement due to a wave moving in the positive x-direction is given by y-11-x2 at time t-0 and by y-1-1-x-12- at t-2 seconds- where x and y are in metres- The velocity of the wave in m-s is-

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Standard XII

Physics

Resultant Wave Equation

The displacem...

Question

The displacement due to a wave moving in the positive $x$ -direction is given by $y = \frac{1}{(1 + x^{2})}$ at time $t = 0$ and by $y = \frac{1}{[1 + (x - 1)^{2}]}$ at $t = 2$ seconds, where $x$ and $y$ are in metres. The velocity of the wave in $m/s$ is:

0.5

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Solution

The correct option is A $0.5$
The general equation of a progressive wave travelling in $+ x$ direction is $y = f (x - v t)$

We rewrite the given equations as $y = \frac{1}{1 + (x - v t)^{2}}$
For $t = 0$ , this becomes $y = \frac{1}{(1 + x^{2})}$
and for $t = 2 s$ , this becomes $y = \frac{1}{[1 + (x - 2 v)^{2}]} = \frac{1}{[1 + (x - 1)^{2}]}$
$\Rightarrow 2 v = 1$ or $v = 0.5 m/s$

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The displacement due to a wave moving in the positive x-direction is given by y=1(1+x2) at time t=0 and by y=1[1+(x−1)2] at t=2 seconds, where x and y are in metres. The velocity of the wave in m/s is:

The correct option is A 0.5The general equation of a progressive wave travelling in +x direction is y=f(x−vt) We rewrite the given equations as y=11+(x−vt)2 For t=0, this becomes y=1(1+x2) and for t=2s, this becomes y=1[1+(x−2v)2]=1[1+(x−1)2] ⇒2v=1 or v=0.5 m/s

The displacement due to a wave moving in the positive $x$ -direction is given by $y = \frac{1}{(1 + x^{2})}$ at time $t = 0$ and by $y = \frac{1}{[1 + (x - 1)^{2}]}$ at $t = 2$ seconds, where $x$ and $y$ are in metres. The velocity of the wave in $m/s$ is: