Question

Which of the following functions correctly represent the travelling wave equation for finite values of displacement and time?

a.y = {x^2} - {t^2}\\\ b.y = \cos {x^2}\sin t\\\ c.y = \log ({x^2} - {t^2}) - \log (x - t)\\\ d.y = {e^{2x}}\sin t \end{array}$$

Answer 1

When attempting questions based on travelling wave equations keep in mind that any and every travelling wave has to satisfy the wave equation and it can be of any form of function as long as it satisfies the wave equation. For attempting this question check each equation for if it satisfies the wave equation or not.
Complete step-by-step solution:
A travelling wave function is represented by a function,
$y = f(x \pm vt)$
For each option we check if the function satisfies the wave equation, if it does then it is the correct function representation.
In option a.
we have
$y = {x^2} - {t^2}$
\Rightarrow $$$$y = (x - t)(x + t) (Equation 1)
Let $(x - t) = {y_1}$
And $(x + t) = {y_2}$
Put in equation 1 , we get
$\therefore y = ({y_1})({y_2})$
Now this equation can’t be a travelling wave because multiplication of two waves is not a travelling wave.
Coming to second option, that is the second equation, we have
$y = \cos {x^2}\sin t$
This equation also doesn’t satisfy our wave equation, firstly because we cannot manipulate or fix this equation into our basic wave equation $y = f(x \pm vt)$and also because here the equation is in standing wave form, its amplitude is varying with$x$. So option b. also won’t satisfy the travelling wave equation because it is a standing wave.
In equation d. $y = {e^{2x}}\sin t$ the amplitude is varying exponentially (${e^{2x}}$), so hence it also does not satisfy our travelling wave equation.
Earlier we said that $y = f(x \pm vt)$ represents the function of a wave, but actually in the wave equation, $y = f(x \pm vt)$ it is not always true that all functions of a wave following this equation represent a wave. For a function to be a wave, it has to follow the wave equation;
\Rightarrow $$$$\dfrac{{{\partial ^2}f}}{{\partial {x^2}}} = \dfrac{1}{{{v^2}}}\dfrac{{{\partial ^2}f}}{{\partial {t^2}}}(Equation 2)
Now let us look at option c. where the equation given to us is as follows,
$y = \log ({x^2} - {t^2}) - \log (x - t)$
\Rightarrow $$$$y = \log \dfrac{{{x^2} - {t^2}}}{{(x - t)}}
\Rightarrow $$$$(\log a - \log b = \log (\dfrac{a}{b}))
\Rightarrow $$$$y = \log (x + t)
\Rightarrow $$$$\dfrac{{\partial y}}{{\partial x}} = \dfrac{1}{{x + t}}
\Rightarrow $$$$\dfrac{{{\partial ^2}y}}{{\partial {x^2}}} = \dfrac{1}{{(x + {t^2})}}
\Rightarrow $$$$\dfrac{{\partial y}}{{\partial t}} = \dfrac{{\partial y}}{{\partial x}}\dfrac{{\partial x}}{{\partial t}}
\Rightarrow $$$$\dfrac{{\dfrac{{\delta x}}{{\delta t}}}}{{x + t}} = \dfrac{v}{{(x + t)}}
\Rightarrow $$$$\dfrac{{{\delta ^2}y}}{{\delta {t^2}}} = \dfrac{{{v^2}}}{{(x + {t^2})}}\dfrac{{{\delta ^2}y}}{{\delta {x^2}}}
$\therefore \dfrac{1}{{{v}^{2}}}\dfrac{{{\delta }^{2}}y}{\delta {{t}^{2}}}$
From equation (2) we can see that equation $y = \log ({x^2} - {t^2}) - \log (x - t)$ follows or satisfies the wave function equation. So we can say that $y = \log ({x^2} - {t^2}) - \log (x - t)$ is correctly representing the travelling wave equation for finite values of $x$ and $t$.
Hence option (c) is the correct answer.

Note: When we talk about waves, we talk about mechanical waves which are governed by all Newton’s laws of motion, or we talk about Electromagnetic waves which are related to electric and magnetic fields and do not need a medium to be propagated. There are other waves too like Matter waves which are associated with moving electrons, protons, neutrons and other fundamental particles etc.