Answer

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**Hint**Any travelling wave must satisfy the wave equation and it can have any form of the function as long as it satisfies the wave equation. We need to check all the three options for whether they satisfy the wave equation or not.

**Complete step by step answer**

A travelling wave function is indeed represented by a function $y = f\left( {x \pm vt} \right)$ but it is not true that all functions of the form $y = f\left( {x \pm vt} \right)$ will represent a wave. For a function to be a wave, it must satisfy the wave equation which is as follows:

$\Rightarrow \dfrac{{{\partial ^2}f}}{{\partial {x^2}}} = \dfrac{1}{{{v^2}}}\dfrac{{{\partial ^2}f}}{{\partial {t^2}}}$

So, the converse statement is not true. Now, let us check all the 3 options and whether they satisfy the wave equation or not.

For option (A),

$\Rightarrow f = {(x - vt)^2}$

We can evaluate the left-hand side as

$\Rightarrow \dfrac{{{\partial ^2}f}}{{\partial {x^2}}} = \dfrac{\partial }{{\partial x}}(2(x - vt))$

$\Rightarrow \dfrac{{{\partial ^2}f}}{{\partial {x^2}}} = 2$

We can evaluate the derivative on the right-hand side as

$\Rightarrow \dfrac{{{\partial ^2}f}}{{\partial {t^2}}} = \dfrac{\partial }{{\partial t}}( - 2v(x - vt))$

$\Rightarrow \dfrac{{{\partial ^2}f}}{{\partial {t^2}}} = 2{v^2}$

Then the right-hand side is

$\Rightarrow \dfrac{1}{{{v^2}}}\dfrac{{{\partial ^2}f}}{{\partial {t^2}}} = \dfrac{{2{v^2}}}{{{v^2}}} = 2$

Since the left-hand side is equal to the right-hand side, the function in option (A) represents a travelling wave.

For option (B),

$\Rightarrow f = log[(x + vt)/{x_0}]$

The left-hand side can be evaluated as:

$\Rightarrow \dfrac{{{\partial ^2}f}}{{\partial {x^2}}} = \dfrac{\partial }{{\partial x}}\left( {\dfrac{{{x_0}}}{{x + vt}}} \right)$

$\Rightarrow \dfrac{{{\partial ^2}f}}{{\partial {x^2}}} = \left( {\dfrac{{ - {x_0}}}{{{{(x + vt)}^2}}}} \right)$

The derivative in the right-hand side can be evaluated as:

$\Rightarrow \dfrac{{{\partial ^2}f}}{{\partial {t^2}}} = v\dfrac{\partial }{{\partial t}}\left( {\dfrac{{{x_0}}}{{x + vt}}} \right)$

$\Rightarrow \dfrac{{{\partial ^2}f}}{{\partial {t^2}}} = {v^2}\left( {\dfrac{{ - {x_0}}}{{{{\left( {x + vt} \right)}^2}}}} \right)$

Then the right-hand side can be calculated as

$\Rightarrow \dfrac{1}{{{v^2}}}\dfrac{{{\partial ^2}f}}{{\partial {t^2}}} = \dfrac{{{v^2}}}{{{v^2}}}\left( {\dfrac{{ - {x_0}}}{{{{\left( {x + vt} \right)}^2}}}} \right)$

$\Rightarrow \dfrac{1}{{{v^2}}}\dfrac{{{\partial ^2}f}}{{\partial {t^2}}} = \left( {\dfrac{{ - {x_0}}}{{{{\left( {x + vt} \right)}^2}}}} \right)$

Since the left-hand side is equal to the right-hand side, the function in option (B) represents a travelling wave.

For option (C),

$\Rightarrow f = 1/(x + vt)$

The left-hand side can be evaluated as:

$\Rightarrow \dfrac{{{\partial ^2}f}}{{\partial {x^2}}} = \dfrac{\partial }{{\partial x}}\left( {\dfrac{{ - 1}}{{{{\left( {x + vt} \right)}^2}}}} \right)$

$\Rightarrow \dfrac{{{\partial ^2}f}}{{\partial {x^2}}} = \dfrac{2}{{{{\left( {x + vt} \right)}^3}}}$

The derivative in the right-hand side can be evaluated as:

$\Rightarrow \dfrac{{{\partial ^2}f}}{{\partial {t^2}}} = v\dfrac{\partial }{{\partial t}}\left( {\dfrac{{ - 1}}{{{{\left( {x + vt} \right)}^2}}}} \right)$

$\Rightarrow \dfrac{{{\partial ^2}f}}{{\partial {t^2}}} = {v^2}\left( {\dfrac{2}{{{{\left( {x + vt} \right)}^3}}}} \right)$

Then the right-hand side can be calculated as

$\Rightarrow \dfrac{1}{{{v^2}}}\dfrac{{{\partial ^2}f}}{{\partial {t^2}}} = \dfrac{{{v^2}}}{{{v^2}}}\left( {\dfrac{2}{{{{\left( {x + vt} \right)}^3}}}} \right)$

$\Rightarrow \dfrac{1}{{{v^2}}}\dfrac{{{\partial ^2}f}}{{\partial {t^2}}} = \dfrac{2}{{{{\left( {x + vt} \right)}^3}}}$

Since the left-hand side is equal to the right-hand side, the function in option (C) represents a travelling wave.

So, all three functions represent travelling waves.

**Option (D) is correct.**

**Note**

While all the three functions given in the question satisfied the wave equation, the converse of the statement that all functions of the form $y = f\left( {x \pm vt} \right)$ will represent a wave is not true as there are a lot of functions that will have the form $y = f\left( {x \pm vt} \right)$ and will still not satisfy the wave equation which is the primary criteria for a function to represent a traveling wave.

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