The following equations represent transverse waves:
z1 = A cos(kx - ωt)
z2 = A cos(kx + ωt)
z3 = A c

Q: Why can't $z_1$ and $z_3$ form a standing wave?

A standing wave requires two identical waves travelling in exactly opposite directions along the same line. Since $z_1$ propagates along the $X$-axis and $z_3$ propagates along the $Y$-axis, they are perpendicular to each other and cannot form a standing wave.

Q: How does the term $\cos\left(\frac{k(x+y)}{2} - \omega t\right)$ represent propagation at $45^\circ$?

The phase of a wave is given by $\vec{k} \cdot \vec{r} - \omega t$. Here, the spatial part is $\frac{k}{2}x + \frac{k}{2}y$, which can be written as $\vec{k} \cdot \vec{r}$ with $\vec{k} = \frac{k}{2}\hat{i} + \frac{k}{2}\hat{j}$. Since the components along $x$ and $y$ are equal, the wave vector points at exactly $45^\circ$ to both axes.

Q: Why is the intensity zero when the amplitude is zero?

The intensity of a wave is directly proportional to the square of its amplitude ($I \propto A^2$). Therefore, at any point where the resultant amplitude is zero, the intensity of the wave will also be zero at all times.

Q: What is the physical meaning of the lines $x - y = (2n + 1)\frac{\pi}{k}$?

These lines represent the 'nodes' of the two-dimensional interference pattern. Along these lines, the two waves $z_1$ and $z_3$ always interfere destructively, meaning the medium along these lines remains completely undisturbed.

Q: Does the frequency of the wave change after superposition?

No, the angular frequency $\omega$ remains unchanged because both combining waves have the exact same frequency. The superposition only redistributes the energy in space, creating regions of maximum and zero intensity.

Question

The following equations represent transverse waves:
$$z_1 = A \cos(kx - \omega t)$$
$$z_2 = A \cos(kx + \omega t)$$
$$z_3 = A \cos(ky - \omega t)$$
Identify the combination(s) of the waves which will produce:
(a) standing wave(s),
(b) a wave travelling in the direction making an angle of $45^\circ$ with the positive $X$ and positive $Y$-axes.
In each case, find the position at which the resultant intensity is always zero.

Sigma Percentile · Accepted Answer

### Introduction to Wave Superposition

Imagine throwing two pebbles into a still pond.

As the ripples expand, they cross paths, creating a beautiful, intricate pattern of peaks and troughs.

This is the **principle of superposition** in action, and it is one of the most fundamental concepts in all of physics.

In this problem, we explore what happens when we superimpose different combinations of three transverse waves propagating in space.

Let's dive deep into the mathematics and physical intuition behind standing waves and directional wave propagation.

---

### Analyzing the Wave Directions

Before writing down any equations, we must first understand the physical direction of each wave.

We are given three wave equations:

$$z_1 = A \cos(kx - \omega t)$$

$$z_2 = A \cos(kx + \omega t)$$

$$z_3 = A \cos(ky - \omega t)$$

Let's look at the phase arguments of these cosine functions.

For $z_1$, the phase is $kx - \omega t$. Since the coefficients of $x$ and $t$ have opposite signs, this wave is propagating in the **positive $X$-direction**.

For $z_2$, the phase is $kx + \omega t$. Since the coefficients of $x$ and $t$ have the same sign, this wave is propagating in the **negative $X$-direction**.

For $z_3$, the phase is $ky - \omega t$. By the same logic, this wave is propagating in the **positive $Y$-direction**.

---

### Part (a): Creating Standing Waves

What is a standing wave?

It is a wave that appears to vibrate in place without propagating through space.

To create a standing wave, we need **two identical waves travelling in opposite directions**.

Looking at our three waves, $z_1$ and $z_2$ fit this description perfectly.

They have the same amplitude $A$, the same angular frequency $\omega$, and the same wavenumber $k$, but they travel in opposite directions along the $X$-axis.

Let's superimpose them:

$$z = z_1 + z_2 = A \cos(kx - \omega t) + A \cos(kx + \omega t)$$

Using the trigonometric identity:

$$\cos C + \cos D = 2 \cos\left(\frac{C+D}{2}ight) \cos\left(\frac{C-D}{2}ight)$$

We can simplify the sum:

$$z = 2A \cos\left(\frac{(kx - \omega t) + (kx + \omega t)}{2}ight) \cos\left(\frac{(kx - \omega t) - (kx + \omega t)}{2}ight)$$

$$z = 2A \cos(kx) \cos(-\omega t)$$

Since $\cos(-	heta) = \cos	heta$, we get:

$$z = 2A \cos(kx) \cos(\omega t)$$

This is the classic equation of a **standing wave**.

Notice how the spatial part $\cos(kx)$ and the temporal part $\cos(\omega t)$ are completely separated!

---

### Finding the Nodes of the Standing Wave

For the resultant intensity to be always zero, the amplitude of the standing wave must be zero at all times.

The amplitude at any position $x$ is given by:

$$A_x = 2A \cos(kx)$$

Setting this amplitude to zero:

$$\cos(kx) = 0$$

We know that the cosine function vanishes at odd multiples of $\frac{\pi}{2}$:

$$kx = (2n + 1)\frac{\pi}{2} \quad 	ext{where } n = 0, \pm 1, \pm 2, \dots$$

Solving for $x$, we find the positions of the **nodes**:

$$x = (2n + 1)\frac{\pi}{2k}$$

At these precise coordinates, the medium remains completely stationary, and the intensity of the wave is always zero.

---

### Part (b): Propagation at $45^\circ$

Now, let's find a combination that produces a wave propagating at $45^\circ$ to both the positive $X$ and positive $Y$ axes.

For a wave to travel in this diagonal direction, its wave vector $\vec{k}$ must have equal components along both axes:

$$\vec{k} = k\hat{i} + k\hat{j}$$

This means the phase of the wave must contain the term $(kx + ky - \omega t)$.

Let's try superimposing $z_1$ (travelling along $+x$) and $z_3$ (travelling along $+y$):

$$z' = z_1 + z_3 = A \cos(kx - \omega t) + A \cos(ky - \omega t)$$

Applying the cosine sum identity again:

$$z' = 2A \cos\left(\frac{(kx - \omega t) + (ky - \omega t)}{2}ight) \cos\left(\frac{(kx - \omega t) - (ky - \omega t)}{2}ight)$$

$$z' = 2A \cos\left(\frac{k(x+y)}{2} - \omega tight) \cos\left(\frac{k(x-y)}{2}ight)$$

Let's analyze this beautiful result!

The term $\cos\left(\frac{k(x+y)}{2} - \omega tight)$ represents a wave propagating along the line $y = x$, which is exactly at **$45^\circ$** to both axes.

The term $\cos\left(\frac{k(x-y)}{2}ight)$ is a purely spatial modulation term that does not change with time.

---

### Finding the Zero Intensity Lines

To find where the resultant intensity is always zero, we look at the spatial amplitude modulation term:

$$A_{x,y} = 2A \cos\left(\frac{k(x-y)}{2}ight)$$

For the intensity to be permanently zero, this amplitude must vanish:

$$\cos\left(\frac{k(x-y)}{2}ight) = 0$$

This occurs when the argument is an odd multiple of $\frac{\pi}{2}$:

$$\frac{k(x-y)}{2} = (2n + 1)\frac{\pi}{2} \quad 	ext{where } n = 0, \pm 1, \pm 2, \dots$$

Cancelling the factor of $2$ from both denominators:

$$k(x-y) = (2n + 1)\pi$$

$$x - y = (2n + 1)\frac{\pi}{k}$$

These equations represent a set of **parallel straight lines** inclined at $45^\circ$ to the axes, along which the wave amplitude is always zero.

This is a stunning demonstration of how simple mathematical identities reveal the deep, elegant structure of physical wave interference!

Visualized Solution

Visualizing the Wave Equations

Condition for Standing Waves

Superposition of $z_{1}$ and $z_{2}$

Resultant Standing Wave Equation

Finding Positions of Zero Intensity (Nodes)

Nodal Positions for Standing Wave

Analyzing Wave Propagation at $4 5^{\circ}$

Superposition of $z_{1}$ and $z_{3}$

Resultant Wave Equation at $4 5^{\circ}$

Finding Positions of Zero Intensity

Zero Intensity Lines

The Way Forward

The Sigma Insight: Interference of Waves

Introduction to Wave Superposition

Analyzing the Wave Directions

Part (a)

Finding the Nodes of the Standing Wave

Part (b)

Finding the Zero Intensity Lines

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