Q.Given below are some functions of $x$ and $t$ to represent the displacement (transverse or longitudinal) of an elastic wave. State which of these represent

Concept: Travelling wave $\Leftrightarrow y=f(x\pm vt)$ — a single, physically valid (finite everywhere, single-valued) function of the combined variable; stationary wave $\Leftrightarrow y=f(x)g(t)$ (separable into a pure function of $x$ times a pure function of $t$).

1. $2\cos(3x)\sin(10t)$: product form $f(x)g(t)$ -- stationary.
2. $2\sqrt{x-vt}$: although it depends only on the combination $(x-vt)$, it is undefined (imaginary) for $x<vt$ and grows without bound as $x-vt\to\infty$ -- it fails the basic physical requirement that a wave's displacement stay finite everywhere at all times. It is also not periodic, so it isn't separable into $f(x)g(t)$ either. It represents neither a travelling wave nor a stationary wave. …

The two diagnostic forms -- and the hidden third requirement

• Travelling wave: $y(x,t)=f(x\pm vt)$ -- the shape moves rigidly through space at speed $v$. But this alone isn't sufficient: a physically acceptable wave function must also remain finite for every value of $x$ and $t$ -- an unbounded or undefined function cannot represent a real physical displacement.
• Stationary wave: $y(x,t)=f(x)\,g(t)$ -- a fixed spatial pattern whose amplitude oscillates in time.

(a) $y=2\cos(3x)\sin(10t)$

This is a pure product of a function of $x$ alone and a function of $t$ alone -- exactly the $f(x)g(t)$ form. This is a stationary wave, with fixed nodes where $\cos(3x)=0$ and antinodes where $|\cos(3x)|=1$.

(b) $y=2\sqrt{x-vt}$

At first glance, $x$ and $t$ appear only in the single combination $(x-vt)$, which looks like the travelling-wave signature. But check whether it's a physically valid wave function:

• For $x<vt$, the quantity under the square root is negative, so $y$ is not even real-valued there.
• For $x-vt\to\infty$, $y=2\sqrt{x-vt}\to\infty$ -- the displacement grows without bound, which no real physical medium can do.

A genuine wave disturbance must be a finite, single-valued function of $(x\pm vt)$ for all $x$ and $t$ -- exactly the requirement this function violates. It is also not periodic, so it cannot be written in the separable $f(x)g(t)$ stationary-wave form either. This function therefore represents neither a travelling wave nor a stationary wave.

(c) $y=3\sin(5x-0.5t)+4\cos(5x-0.5t)$

Both terms share the identical argument $\theta=5x-0.5t$. Using $A\sin\theta+B\cos\theta=R\sin(\theta+\alpha)$ with $R=\sqrt{A^2+B^2}$:

$R = \sqrt{3^2+4^2} = 5, \qquad \tan\alpha = \frac43$ …

Step 1: Recall the two diagnostic forms: a travelling wave must be a finite, physically valid function of the single combination $(x\pm vt)$; a stationary wave must separate into $f(x)\cdot g(t)$.

Step 2: (a) $y=2\cos(3x)\sin(10t)$ is exactly product-separable $\Rightarrow$ stationary.

Step 3: (b) $y=2\sqrt{x-vt}$ depends only on $(x-vt)$, but it is undefined (imaginary) for $x<vt$ and grows without bound as $x-vt\to\infty$ — it fails the requirement that a wave's displacement stay finite everywhere, so it does not represent a travelling wave; it's also not periodic/separable, so not stationary either — it represents neither.

Step 4: (c) $y=3\sin(5x-0.5t)+4\cos(5x-0.5t)$: both terms share the identical argument $\theta=5x-0.5t$; combine with $A\sin\theta+B\cos\theta=R\sin(\theta+\alpha)$, $R=\sqrt{3^2+4^2}=5$, giving $y=5\sin(5x-0.5t+\alpha)$ — a single function of $(x-vt)$ $\Rightarrow$ travelling, $v=\omega/k=0.5/5=0.1\ \text{m/s}$. …