Q.Given below are some functions of $x$ and $t$ to represent the displacement of an elastic wave.

The key idea is to classify each wave form by its mathematical structure: a travelling wave depends on $x \pm vt$ (or $kx \pm \omega t$), a stationary wave is a product of separate functions of $x$ and $t$, and beats arise from the sum of two close frequencies.

(a) $y = 5\cos(4x)\sin(20t)$ is a product of a function of $x$ alone and a function of $t$ alone — this is the standard form of a stationary wave.

(b) $y = 4\sin(5x - t/2) + 3\cos(5x - t/2)$ is a sum of two terms, each depending on the same combination $(5x - t/2)$. This is a travelling wave along $+x$ (since the sign of the $t$ term is negative). …

Why This Approach Works

Every elastic wave is described by a function $y(x,t)$. The classification hinges on how $x$ and $t$ appear together:

• A travelling wave has the form $f(kx \pm \omega t + \phi)$ — the variables are locked in a single combination. The sign tells the direction: $kx - \omega t$ means $+x$ direction, $kx + \omega t$ means $-x$ direction.
• A stationary (standing) wave is a product $f(x) \cdot g(t)$ — the $x$ and $t$ dependences are separate, so the wave doesn't propagate; it oscillates in place.
• Beats arise from the superposition of two waves with slightly different frequencies. The product of two cosines with different arguments, when expanded, gives a sum of two travelling waves with close frequencies — but the given form $y = 10\cos[(252-250)\pi t]\cos[(252+250)\pi t]$ is already the product form that represents the envelope (beat) and carrier.

Let's examine each case.

1. Function (a): $y = 5\cos(4x)\sin(20t)$

   This is a pure product: a function of $x$ alone multiplied by a function of $t$ alone. There is no combination like $kx \pm \omega t$. This is the classic form of a stationary wave — nodes and antinodes are fixed in space, and every point oscillates in phase (or opposite phase) with a time-dependent amplitude.

   Watch out

   Don't be fooled by the sine and cosine being different — the key is the separation of variables, not the trigonometric function used. $\cos(4x)\sin(20t)$ is just as much a standing wave as $\sin(4x)\cos(20t)$.

   Conclusion: Stationary wave.

2. Function (b): $y = 4\sin(5x - t/2) + 3\cos(5x - t/2)$

   Both terms contain the same combination $(5x - t/2)$. This is a sum of two travelling waves with the same speed and direction. Using the identity $R\sin(\theta + \phi) = A\sin\theta + B\cos\theta$, we can combine them into a single sine wave:

where $R = \sqrt{4^2 + 3^2} = 5$ and $\phi = \tan^{-1}(3/4)$. The argument is $5x - t/2$, which is of the form $kx - \omega t$ — so this is a travelling wave along $+x$ direction.

Conclusion: Travelling wave along $+x$.

3. Function (c): $y = 10\cos[(252 - 250)\pi t]\cos[(252+250)\pi t]$

   Simplify the arguments:

This is a product of two cosines with different frequencies. Using the identity $\cos A \cos B = \frac{1}{2}[\cos(A+B) + \cos(A-B)]$, we get:

$y = 5[\cos(504\pi t) + \cos(500\pi t)]$ …

Step 1: Recall the diagnostic forms: a travelling wave depends on a single combination $(kx\pm\omega t)$; a stationary wave is a product of a function of $x$ alone and a function of $t$ alone; beats show up as a slow-envelope × fast-carrier product with no leftover spatial ($x$) dependence in the given expression.

Step 2 — (a) $y=5\cos(4x)\sin(20t)$: This is already a pure product $f(x)\cdot g(t)$, with no combined argument $kx\pm\omega t$ anywhere — the signature of a stationary wave.

Step 3 — (b) $y=4\sin(5x-t/2)+3\cos(5x-t/2)$: Both terms share the identical argument $(5x-t/2)$. Combine via $A\sin\theta+B\cos\theta=R\sin(\theta+\phi)$ to get a single wave $R\sin(5x-t/2+\phi)$. The form $kx-\omega t$ (with $k=5$, $\omega=1/2$) is a travelling wave along $+x$ (constant phase requires $x$ to increase as $t$ increases). …