A plane progressive wave of frequency 25 Hz, amplitude 2.5 x 10^-5 m and initial phase zero propagat

Q: Why does the direction of propagation (negative x-direction) not affect the phase difference calculation?

The phase difference depends only on the spatial separation (distance) between the two points along the line of propagation, not on the direction in which the wave is traveling. The distance $\Delta x = 6\text{ m}$ remains positive regardless of the wave's direction.

Q: What would happen to the amplitude difference if the medium was absorbing?

If the medium absorbed energy, the amplitude would decay exponentially with distance ($A(x) = A_0 e^{-\alpha x}$). In that case, the amplitude at the second point would be smaller, resulting in a non-zero amplitude difference.

Q: Is phase difference always measured in radians?

In standard physics equations, phase difference is calculated in radians. However, it can also be expressed in degrees. For example, $\pi$ radians is equivalent to $180^\circ$.

Q: How does a plane wave differ from a spherical wave regarding amplitude?

For a plane wave, amplitude remains constant with distance ($A = \text{constant}$). For a spherical wave originating from a point source, the amplitude decreases inversely with distance ($A \propto \frac{1}{r}$) because the wave energy spreads over a sphere of area $4\pi r^2$.

Q: What does a phase difference of $2\pi$ mean physically?

A phase difference of $2\pi$ radians (or $360^\circ$) means the two points are in phase. They oscillate in perfect synchronization, reaching their crests and troughs at the exact same times.

Question

A plane progressive wave of frequency $25	ext{ Hz}$, amplitude $2.5 	imes 10^{-5}	ext{ m}$ and initial phase zero propagates along the negative $x$-direction with a velocity of $300	ext{ m/s}$. At any instant, the phase difference between the oscillations at two points $6	ext{ m}$ apart along the line of propagation is ...... and the corresponding amplitude difference is ...... m.

Sigma Percentile · Accepted Answer

### Introduction to Wave Propagation

Waves are one of the most beautiful and fundamental phenomena in physics. From the ripples on a quiet pond to the light traveling across the vastness of space, waves carry energy and information from one place to another without transferring matter.

In this problem, we dive deep into the kinematics of a **plane progressive wave**. We are tasked with finding two key quantities: the **phase difference** and the **amplitude difference** between two points separated by a distance of $6	ext{ m}$ along the line of propagation.

Let's break down the physics step-by-step and see how beautifully the math aligns with physical reality.

---

### Step 1: Finding the Spatial Period (Wavelength)

Before we can talk about phase differences, we need to understand the spatial scale of our wave. Just as a wave has a temporal period $T$ (the time it takes for one complete oscillation at a single point), it also has a spatial period $\lambda$, known as the **wavelength**.

The wavelength is the distance over which the wave's shape repeats itself in space. We are given:
- Frequency of the wave, $f = 25	ext{ Hz}$
- Velocity of the wave, $v = 300	ext{ m/s}$

These quantities are bound together by the fundamental wave speed equation:

$$v = f \lambda$$

Rearranging this equation to solve for the wavelength $\lambda$, we get:

$$\lambda = \frac{v}{f}$$

Substituting the given values:

$$\lambda = \frac{300	ext{ m/s}}{25	ext{ Hz}} = 12	ext{ m}$$

This tells us that one complete cycle of our wave spans exactly $12	ext{ m}$ in space.

---

### Step 2: Calculating the Phase Difference

Now, let's think about phase. Phase represents the position of a point in time or space on a wave cycle. A full cycle corresponds to a phase of $2\pi$ radians (or $360^\circ$).

If a full cycle of $12	ext{ m}$ corresponds to a phase change of $2\pi$ radians, then any spatial separation $\Delta x$ (path difference) will correspond to a proportional phase difference $\Delta \phi$:

$$\Delta \phi = \frac{2\pi}{\lambda} \Delta x$$

We are interested in two points separated by $\Delta x = 6	ext{ m}$. Let's substitute our values:

$$\Delta \phi = \frac{2\pi}{12	ext{ m}} 	imes 6	ext{ m}$$

$$\Delta \phi = \pi	ext{ radians}$$

#### Physical Interpretation of $\pi$ Phase Difference

A phase difference of $\pi$ radians (or $180^\circ$) has a profound physical meaning. It means the two points are in **opposite phase** (completely out of phase).

When the particle at the first point is at its maximum positive displacement (a crest), the particle at the second point is at its maximum negative displacement (a trough). They are moving in opposite directions at any given instant, perfectly mirroring each other's motion across the equilibrium line.

---

### Step 3: Analyzing the Amplitude Difference

The second part of the question asks for the amplitude difference between these two points.

To answer this, we must look closely at the type of wave described: a **plane progressive wave** propagating in a non-absorbing medium.

In a plane wave, the wavefronts are parallel planes. Unlike spherical waves (where energy spreads out over an increasing spherical surface area), the energy of a plane wave remains confined to a constant cross-sectional area as it propagates.

Because there is no spreading of energy and we assume no absorption by the medium, the energy density—and consequently, the **amplitude** of the wave—remains absolutely constant at all points along the direction of propagation.

Therefore:

$$A_1 = A_2 = 2.5 	imes 10^{-5}	ext{ m}$$

The amplitude difference $\Delta A$ is:

$$\Delta A = A_1 - A_2 = 0	ext{ m}$$

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### Summary of Results

- **Phase Difference:** $\pi$ radians
- **Amplitude Difference:** $0	ext{ m}$

This elegant result highlights the clean, symmetric nature of plane waves in ideal media. It serves as a foundational building block for understanding wave interference, standing waves, and diffraction!

Visualized Solution

Understanding the Wave Parameters

The Wavelength Formula

Substituting the Values

Calculating Wavelength

Phase Difference Formula

Substituting Path Difference

Calculating Phase Difference

Analyzing Amplitude Difference

Calculating Amplitude Difference

The Way Forward

The Sigma Insight: Wave Equation and Wave Speed

Introduction to Wave Propagation

Step 1

Step 2

# Physical Interpretation of $π$ Phase Difference

Step 3

Summary of Results

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