When two progressive waves y1 = 4 sin(2x - 6t) and y2 = 3 sin(2x - 6t - pi/2) are superimposed, the

Q: Why can we treat wave amplitudes like vectors?

Because harmonic waves can be represented as phasors (rotating vectors) in a complex plane. Since they share the same frequency, the phase difference between them remains constant, allowing us to use vector addition to find the resultant amplitude.

Q: Does the phase difference of $-\pi/2$ mean the amplitude is negative?

No, amplitude is always a positive scalar quantity representing the maximum displacement. The negative sign in the phase $-\pi/2$ simply indicates a phase lag, which determines the direction of the phasor vector, not the magnitude of the amplitude.

Q: What if the frequencies of the two waves were different?

If the frequencies were different, the phase difference would change continuously over time. In that case, we would observe beats rather than a constant resultant amplitude, and the simple phasor addition method would not apply directly.

Q: Is the resultant wave also a progressive wave?

Yes! The superposition of two progressive waves of the same frequency and wavelength travelling in the same direction results in another progressive wave of the same frequency and wavelength, but with a modified amplitude and phase.

Q: What is the phase of the resultant wave?

Using the phasor diagram, the resultant vector makes an angle $\theta$ with the horizontal vector $\vec{A}_1$. We can find this angle using $\tan\theta = A_2 / A_1 = 3/4$, which gives $\theta \approx -37^\circ$ or $-0.64$ radians. Thus, the resultant wave equation is $y = 5 \sin(2x - 6t - 0.64)$.

Question

When two progressive waves $y_1 = 4 \sin (2x - 6t)$ and $y_2 = 3 \sin \left(2x - 6t - \frac{\pi}{2}\right)$ are superimposed, the amplitude of the resultant wave is

Sigma Percentile · Accepted Answer

### Introduction to Wave Superposition

Imagine two ripples on a calm pond racing toward each other.

When they collide, they don't bounce off like billiard balls.

Instead, they pass right through each other, momentarily merging into a single, complex shape before continuing on their separate ways.

This beautiful phenomenon is governed by the **Principle of Superposition**, one of the most fundamental laws of wave mechanics.

In this problem, we are asked to find the resultant amplitude when two progressive harmonic waves, $y_1$ and $y_2$, superimpose in the same medium.

Let's dive deep into the mathematics and geometry behind this elegant process!

### Analyzing the Wave Equations

Let's write down the equations of the two waves given to us:

$$y_1 = 4 \sin(2x - 6t)$$

$$y_2 = 3 \sin\left(2x - 6t - \frac{\pi}{2}ight)$$

To understand how these waves interact, we must first compare them to the standard equation of a progressive harmonic wave:

$$y = A \sin(kx - \omega t + \phi)$$

Here, $A$ represents the **amplitude** (the maximum displacement from equilibrium), $k$ is the wave number, $\omega$ is the angular frequency, and $\phi$ is the initial phase.

By comparing our two waves to this standard form, we can extract their individual amplitudes:

For the first wave, the amplitude is:

$$A_1 = 4$$

For the second wave, the amplitude is:

$$A_2 = 3$$

Notice that both waves have the exact same wave number ($k = 2 	ext{ rad/m}$) and the same angular frequency ($\omega = 6 	ext{ rad/s}$).

This means they are **coherent sources**—they have the same wavelength and frequency, which is a crucial requirement for producing a stable interference pattern!

### The Concept of Phase Difference

Now, let's look at the phase of each wave.

The phase of the first wave is $	heta_1 = 2x - 6t$.

The phase of the second wave is $	heta_2 = 2x - 6t - \frac{\pi}{2}$.

The **phase difference** ($\phi$) between the two waves is the difference between their phases:

$$\phi = 	heta_1 - 	heta_2 = (2x - 6t) - \left(2x - 6t - \frac{\pi}{2}ight) = \frac{\pi}{2} 	ext{ radians}$$

A phase difference of $\frac{\pi}{2}$ radians is equivalent to $90^\circ$.

This means that the second wave lags behind the first wave by a quarter of a cycle.

When the first wave is at its maximum displacement, the second wave is just passing through its equilibrium position!

### The Phasor Method: Geometry Meets Waves

How do we add these two waves together?

We could use trigonometric identities, but there is a much more intuitive and elegant method: **Phasors**.

A phasor is a rotating vector whose length represents the amplitude of the wave, and whose angle with the horizontal axis represents the phase of the wave.

Since both waves have the same frequency, their phasors rotate at the same speed, meaning the angle between them remains constant at $\phi = 90^\circ$.

Therefore, we can treat the addition of these two waves exactly like the addition of two vectors!

The first wave is represented by a vector $\vec{A}_1$ of length $4$ along the horizontal axis.

The second wave is represented by a vector $\vec{A}_2$ of length $3$ pointing vertically downwards (due to the $-90^\circ$ phase lag).

The resultant wave is represented by the vector sum:

$$\vec{A} = \vec{A}_1 + \vec{A}_2$$

Since the two vectors are perpendicular, they form a right-angled triangle!

### Calculating the Resultant Amplitude

To find the magnitude of the resultant vector (which is the resultant amplitude $A$), we can use the general vector addition formula:

$$A = \sqrt{A_1^2 + A_2^2 + 2A_1A_2 \cos\phi}$$

Let's substitute our known values into this formula:

$$A = \sqrt{4^2 + 3^2 + 2(4)(3) \cos\left(\frac{\pi}{2}ight)}$$

Since $\cos\left(\frac{\pi}{2}ight) = \cos(90^\circ) = 0$, the third term under the square root becomes zero:

$$A = \sqrt{16 + 9 + 0}$$

$$A = \sqrt{25}$$

$$A = 5$$

The resultant amplitude is exactly **5**.

This is a classic **Pythagorean triple** ($3, 4, 5$), which beautifully illustrates how wave superposition can be simplified into pure geometry!

### Exploring the Extremes of Interference

To truly master this concept, let's look at what would happen if the phase difference $\phi$ were different:

1. **Constructive Interference ($\phi = 0$):**
   If the waves were perfectly in phase, they would reinforce each other fully. The resultant amplitude would be:
   $$A_{max} = A_1 + A_2 = 4 + 3 = 7$$

2. **Destructive Interference ($\phi = \pi$):**
   If the waves were completely out of phase ($180^\circ$), they would oppose each other. The resultant amplitude would be:
   $$A_{min} = |A_1 - A_2| = |4 - 3| = 1$$

Our resultant amplitude of $5$ lies comfortably between these two extreme limits ($1 \le A \le 7$), which perfectly matches our physical intuition!

Animated Solution for Physics - Waves: When two progressive waves $y_{1} = 4 sin (2 x - 6 t)$ and $y_{2} = 3 sin (2 x - 6 t - \frac{π}{2})$ are superimposed, the amplitude of the resultant wave is

Enter Numerical Value:

Visualized Solution

Visualizing the Wave Equations

The Principle of Superposition

Identifying Amplitudes and Phase Difference

The Vector Addition Formula for Phasors

Substituting Parameters into the Formula

Evaluating the Cosine Term

Calculating the Resultant Amplitude

Exploring Other Phase Differences

The Sigma Insight: Interference of Waves

Introduction to Wave Superposition

Analyzing the Wave Equations

The Concept of Phase Difference

The Phasor Method

Calculating the Resultant Amplitude

Exploring the Extremes of Interference

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Visualized Solution

Visualizing the Wave Equations

The Principle of Superposition

Identifying Amplitudes and Phase Difference

The Vector Addition Formula for Phasors

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