You have learnt that a travelling wave in one dimension is represented by a function $y = f(x, t)$ where $x$ and $t$ must appear in the combination $x - vt$ or $x + vt$, i.e. $y = f(x \pm vt)$. Is the converse true? Examine if the following functions for $y$ can possibly represen

Question

You have learnt that a travelling wave in one dimension is represented by a function $y = f(x, t)$ where $x$ and $t$ must appear in the combination $x - vt$ or $x + vt$, i.e. $y = f(x \pm vt)$. Is the converse true? Examine if the following functions for $y$ can possibly represent a travelling wave:
(a) $(x - vt)^{2}$
(b) $\log\left[(x + vt)/x_{0}\right]$
(c) $1/(x + vt)$

Accepted Answer

> [!TLDR] > The statement "$y=f(x,t)$ is a travelling wave only if $x,t$ appear as $x\pm vt$" is a *necessary* condition, not a sufficient one -- the converse is **not simply true**: a function of $(x\pm vt)$ only represents a genuine travelling wave if it stays **finite for all $x$ and $t$** (a real physical disturbance can't have infinite or undefined displacement anywhere). Checking all three functions against this finiteness test, **none** of (a), > (b), > (c) represents a valid travelling wave. ## Why finiteness is the real test A physical wave disturbance $y(x,t)$ must be bounded everywhere and at all times -- a guitar string, a water surface, or an electromagnetic field cannot have an infinite or undefined displacement at some point in space or as time goes on. So even though every function of the pure combination $(x\pm vt)$ *looks* like a travelling wave, we must additionally check that it never blows up or becomes undefined for any real $x,t$. ## (a) $y=(x-vt)^2$ This is of the form $f(x-vt)$ with $f(u)=u^2$, so at any *fixed* pair $(x,t)$ it gives a finite, well-defined value. But as $t o\infty$ (or $x o\infty$), $y=(x-vt)^2 o\infty$ -- a real wave's displacement cannot grow without bound as it propagates. Applying the same disqualifying finiteness test used for (b) and (c), (a) also fails to represent a physically valid travelling wave. ## (b) $y=\log\!\left[\dfrac{x+vt}{x_0} ight]$ This is of the form $f(x+vt)$. But $\log(u) o-\infty$ as $u o0^+$, i.e. as $x+vt o0$, and $\log(u)$ is undefined for $u<0$, i.e. whenever $x+vt<0$. The function diverges at one point and is undefined over half of all $(x,t)$ space. ## (c) $y=\dfrac{1}{x+vt}$ Also of the form $f(x+vt)$, but it diverges to infinity exactly at $x+vt=0$. A function that blows up at any point in its domain cannot represent a real physical wave displacement there. ## Putting it together …

Concept understanding — Wave Type Classification

Wave Type Classification: From Intuition to Precision

The Core Intuition

The Precise Classification

Transverse Waves

Longitudinal Waves

Why This Matters

Why finiteness is the real test

(a) $y=(x-vt)^2$

(b) $y=\log\!\left[\dfrac{x+vt}{x_0}\right]$

(c) $y=\dfrac{1}{x+vt}$

Putting it together …

Concept and Intuition

Step-by-Step Solution

Concept and Intuition

Step-by-Step Solution

Concept and Intuition

Step-by-Step Solution

Concept and Intuition

Step-by-Step Solution …

Concept and Intuition

Step-by-Step Solution

Concept and Intuition

Step-by-Step Solution

Concept and Intuition

Step-by-Step Solution

Concept and Intuition

Step-by-Step Solution

Concept and Intuition

Step-by-Step Solution

Concept and Intuition

Step-by-Step Solution

Why finiteness is the real test

(a) y=(x−vt)2y=(x-vt)^2y=(x−vt)2

(b) y=log⁡ ⁣[x+vtx0]y=\log\!\left[\dfrac{x+vt}{x_0}\right]y=log[x0​x+vt​]

(c) y=1x+vty=\dfrac{1}{x+vt}y=x+vt1​

Putting it together …

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(a) $y=(x-vt)^2$

(b) $y=\log\!\left[\dfrac{x+vt}{x_0}\right]$

(c) $y=\dfrac{1}{x+vt}$

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