Sound waves of frequency 660 Hz fall normally on a perfectly reflecting wall. The shortest distance

Q: Why is the displacement of air particles zero at the wall?

Because the wall is rigid and impenetrable, the air molecules in direct contact with it cannot move physically, forcing their displacement to be zero at all times.

Q: Is a displacement node always a pressure antinode?

Yes! In sound waves, displacement and pressure are $90^\circ$ out of phase. Where particles don't move (node), the compression and rarefaction are maximum, leading to maximum pressure variation (antinode).

Q: Why does reflection at a rigid wall cause a phase change of $\pi$ for displacement?

To satisfy the boundary condition of zero displacement at the wall, the reflected wave must interfere destructively with the incident wave at the boundary, which requires a phase shift of $\pi$ (or $180^\circ$).

Q: What would happen if the wall was not perfectly reflecting?

If the wall absorbed some energy, the reflected wave would have a smaller amplitude. A standing wave would still form, but the nodes would not have completely zero amplitude; they would be partial nodes.

Q: How does this change if the sound wave is incident at an angle instead of normally?

If incident at an angle $\theta$, only the normal component of the wave vector $k_x = k \cos\theta$ contributes to the standing wave pattern perpendicular to the wall, shifting the positions of the nodes and antinodes further away from the wall.

Question

Sound waves of frequency $660	ext{ Hz}$ fall normally on a perfectly reflecting wall. The shortest distance from the wall at which the air particles have maximum amplitude of vibration is ...... $	ext{m}$. Speed of sound $= 330	ext{ m/s}$.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine standing in front of a massive, rigid concrete wall. If you emit a sound wave normally towards it, the wave travels through the air, hits the wall, and bounces straight back.

This physical scenario is a classic example of wave reflection at a boundary. When the incident wave and the reflected wave travel in opposite directions through the same medium, they superimpose.

This superposition leads to the formation of a **standing wave**. Unlike progressive waves, standing waves do not transfer energy through space; instead, they form stationary patterns of oscillation with fixed points of zero motion and maximum motion.

### The Boundary Condition

To solve this problem, we must first establish the boundary condition at the reflecting wall.

Because the wall is rigid and impenetrable, air molecules in direct contact with the wall are physically blocked from moving. They cannot vibrate back and forth.

Therefore, the displacement of air particles at the wall is strictly zero at all times:

$$s(x=0, t) = 0$$

In wave mechanics, any point where the medium's displacement is permanently zero is called a **displacement node**. Thus, the reflecting wall acts as a displacement node in our standing wave pattern.

### Finding the First Antinode

The question asks for the shortest distance from the wall at which the air particles have the **maximum amplitude of vibration**.

Points of maximum vibration amplitude are called **displacement antinodes**.

In any standing wave pattern, nodes and antinodes alternate at regular intervals. The distance between two consecutive nodes is half a wavelength ($\frac{\lambda}{2}$), and the distance between a node and its nearest adjacent antinode is exactly a quarter of a wavelength:

$$d = \frac{\lambda}{4}$$

Since the wall itself is a node, the very first point of maximum vibration (the first antinode) will occur at this shortest distance of $\frac{\lambda}{4}$ from the wall.

### Calculating the Wavelength

To find this distance, we must first calculate the wavelength ($\lambda$) of the sound wave. We are given:
- Frequency of the sound wave, $f = 660	ext{ Hz}$
- Speed of sound in air, $v = 330	ext{ m/s}$

Using the fundamental wave relation:

$$v = f \lambda$$

We can rearrange this to solve for $\lambda$:

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

Substituting the given values:

$$\lambda = \frac{330	ext{ m/s}}{660	ext{ Hz}} = 0.5	ext{ m}$$

### Final Calculation

Now that we have determined the wavelength to be $0.5	ext{ m}$, we can calculate the shortest distance ($d$) to the first displacement antinode:

$$d = \frac{\lambda}{4} = \frac{0.5	ext{ m}}{4} = 0.125	ext{ m}$$

Thus, the shortest distance from the wall at which the air particles vibrate with maximum amplitude is **$0.125	ext{ m}$**.

### A Crucial Distinction: Displacement vs. Pressure

For competitive exams like JEE, it is vital to understand the difference between **displacement waves** and **pressure waves** in acoustics.

While the air particles cannot move at the wall (making it a displacement node), they are constantly being compressed against and rarefied away from the wall. This means that the pressure variations are at their absolute maximum at the wall.

Therefore, a rigid reflecting wall is a **displacement node** but a **pressure antinode**. Always read the question carefully to see whether it asks about particle displacement or pressure variations!

Animated Solution for Physics - Waves: Sound waves of frequency $660 Hz$ fall normally on a perfectly reflecting wall. The shortest distance from the wall at which the air particles have maximum amplitude of vibration is ...... $m$ . Speed of sound $= 330 m/s$ .

Enter Numerical Value:

Visualized Solution

Visualizing the Physical Setup

Applying the Boundary Condition

Identifying the Target Point

Relating Distance to Wavelength

The Wave Relation Formula

Substituting Known Values

Calculating the Wavelength

Calculating the Shortest Distance

Final Answer Evaluation

The Way Forward: Displacement vs. Pressure

The Sigma Insight: Standing Waves in Strings and Organ Pipes

Analyzing the Setup

The Boundary Condition

Finding the First Antinode

Calculating the Wavelength

Final Calculation

A Crucial Distinction

Similar Questions

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Enter Numerical Value:

Visualized Solution

Visualizing the Physical Setup

Applying the Boundary Condition

Identifying the Target Point

Relating Distance to Wavelength

The Wave Relation Formula

Substituting Known Values

Calculating the Wavelength

Calculating the Shortest Distance

Final Answer Evaluation

The Way Forward: Displacement vs. Pressure

The Sigma Insight: Standing Waves in Strings and Organ Pipes

Analyzing the Setup

The Boundary Condition

Finding the First Antinode

Calculating the Wavelength

Final Calculation

A Crucial Distinction

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