Stationary waves

1. Overview

Stationary waves (also known as standing waves) represent a unique wave phenomenon where energy is stored within a medium rather than being transferred from one location to another. They are formed by the superposition of two progressive waves that possess the same frequency (and wavelength) and similar amplitude, but are travelling in opposite directions.

Unlike progressive waves, where every particle in the medium eventually undergoes the same maximum displacement (amplitude), particles in a stationary wave have varying amplitudes depending on their position. This results in a fixed pattern of nodes (points of zero displacement) and antinodes (points of maximum displacement). This topic is fundamental to understanding the physics of musical instruments, microwave technology, and the resonance of physical structures.

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Key Definitions

• Principle of Superposition: When two or more waves meet at a point, the resultant displacement is the algebraic (vector) sum of the displacements of the individual waves at that point.
• Stationary Wave: A wave pattern produced by the superposition of two progressive waves of the same frequency and amplitude travelling in opposite directions. It is characterized by nodes and antinodes and does not transfer energy.
• Node: A point on a stationary wave where the amplitude is permanently zero due to total destructive interference.
• Antinode: A point on a stationary wave where the amplitude is maximum due to constructive interference.
• Coherence: Two waves are coherent if they have a constant phase difference. For a stationary wave to be stable, the two superposing waves must be coherent (which is guaranteed if they have the same frequency).
• Fundamental Frequency ($f_1$ or $f_0$): The lowest frequency at which a stationary wave can be formed in a given system (also called the first harmonic).
• Harmonic: A mode of vibration where the frequency is an integer multiple of the fundamental frequency.

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Content

3.1 The Principle of Superposition

The Principle of Superposition is the governing law for all wave interactions. When waves overlap, they do not "collide" or "bounce"; they pass through one another. During the time they occupy the same space, the medium responds to both simultaneously.

• Constructive Interference: Occurs when the phase difference between two waves is $0$, $2\pi$, or any integer multiple of $2\pi$ radians ($0^\circ, 360^\circ$). The waves are "in phase," and their displacements add to create a maximum resultant displacement.
• Destructive Interference: Occurs when the phase difference is $\pi, 3\pi$, or any odd integer multiple of $\pi$ radians ($180^\circ, 540^\circ$). The waves are "in antiphase," and their displacements subtract. If the amplitudes are equal, the resultant displacement is zero.

3.2 Formation of Stationary Waves: Graphical Method

To understand how a stationary wave forms, we analyze two progressive waves, $W_1$ (moving right) and $W_2$ (moving left), at different time intervals relative to their period $T$.

1. At $t = 0$: $W_1$ and $W_2$ are perfectly in phase. Every peak of $W_1$ aligns with a peak of $W_2$. The resultant wave (the sum) shows maximum possible displacement at specific points. These points of maximum displacement are the Antinodes.
2. At $t = T/4$: Each wave has moved a distance of $\frac{1}{4}\lambda$ in its respective direction. This creates a total relative shift of $\frac{1}{2}\lambda$. The waves are now $180^\circ$ out of phase (antiphase). At every point along the medium, the positive displacement of $W_1$ is cancelled by the negative displacement of $W_2$. The resultant displacement is zero everywhere.
3. At $t = T/2$: Each wave has moved a total distance of $\frac{1}{2}\lambda$. They are back in phase, but both have inverted their positions compared to $t=0$. Peaks now align where troughs were previously. The resultant wave again shows maximum displacement at the same Antinode positions, but in the opposite direction.
4. Nodes: At certain fixed positions, the two waves are always in antiphase, regardless of the time $t$. At these points, the resultant displacement is permanently zero.

The Vibration Envelope: If you were to take a long-exposure photograph of a vibrating string, you would see a "loop" shape. This is the envelope of the stationary wave, showing the limits of the medium's displacement.

3.3 Nodes, Antinodes, and Wavelength

The geometry of a stationary wave is strictly defined by the wavelength $\lambda$ of the original progressive waves:

• Distance between two adjacent nodes (N to N): $\frac{1}{2}\lambda$
• Distance between two adjacent antinodes (A to A): $\frac{1}{2}\lambda$
• Distance between a node and the adjacent antinode (N to A): $\frac{1}{4}\lambda$
• Distance for one full "loop": $\frac{1}{2}\lambda$ (A stationary wave consists of two loops per wavelength).

3.4 Comparison: Progressive vs. Stationary Waves

Feature	Progressive Wave	Stationary Wave
Energy Transfer	Transfers energy in the direction of travel.	No net transfer of energy; energy is stored.
Amplitude	All points have the same amplitude.	Amplitude varies from zero (nodes) to max (antinodes).
Phase	Phase changes continuously along the wave.	All points between two nodes are in phase.
Wavelength	Distance between two adjacent points in phase.	Twice the distance between adjacent nodes.
Frequency	All points vibrate at the same frequency.	All points vibrate at the same frequency (except nodes).

3.5 Experimental Demonstrations

A. Microwaves (Reflection Method)

• Setup: A microwave transmitter emits waves toward a vertical metal reflecting sheet. The reflected wave travels back toward the transmitter, superposing with the incident wave.
• Procedure: A microwave probe (detector) is moved slowly along the line between the transmitter and the reflector.
• Observation: The detector's meter fluctuates between high readings (antinodes) and zero/minimum readings (nodes).
• Measurement: To find the wavelength $\lambda$, measure the distance $d$ across several nodes (e.g., 10 nodes). The distance between two adjacent nodes is $\frac{\lambda}{2}$. Therefore, $\lambda = \frac{2 \times \text{Total Distance}}{\text{Number of gaps between nodes}}$.

B. Stretched Strings (Melde’s Experiment)

• Setup: One end of a string is attached to a vibration generator; the other passes over a pulley to a hanging mass (providing tension $T$).
• Procedure: The frequency $f$ of the generator is adjusted until a stable stationary wave pattern (loops) is seen.
• Boundary Conditions: The end at the pulley and the end at the generator are effectively nodes (fixed points).
• Harmonics:
  • 1st Harmonic (Fundamental): 1 loop. Length $L = \frac{1}{2}\lambda \Rightarrow \lambda = 2L$.
  • 2nd Harmonic: 2 loops. Length $L = \lambda \Rightarrow \lambda = L$.
  • 3rd Harmonic: 3 loops. Length $L = \frac{3}{2}\lambda \Rightarrow \lambda = \frac{2}{3}L$.

C. Air Columns (Sound Waves)

• Setup: A tuning fork of frequency $f$ is struck and held over a tube. The length of the air column $L$ is varied (e.g., by moving a piston or changing the water level).
• Boundary Conditions:
  • Closed end: Air molecules cannot move; must be a node.
  • Open end: Air molecules can move with maximum freedom; must be an antinode.
• Resonance in a Closed Pipe:
  • 1st Resonance: $L = \frac{1}{4}\lambda$ (Node at bottom, Antinode at top).
  • 2nd Resonance: $L = \frac{3}{4}\lambda$.
  • 3rd Resonance: $L = \frac{5}{4}\lambda$.
• Note: Only odd harmonics ($f_1, 3f_1, 5f_1...$) are possible in a pipe closed at one end.

3.6 Worked Examples

Worked example 1 — Microwaves and Frequency

A microwave transmitter is placed $50.0\text{ cm}$ from a metal reflector. A probe is moved from the reflector toward the transmitter. The first minimum is found at the reflector surface, and the sixth minimum is found at a distance of $12.5\text{ cm}$ from the reflector. Calculate the frequency of the microwaves. (Speed of light $c = 3.00 \times 10^8\text{ m s}^{-1}$).

Step 1: Determine the number of node-to-node gaps The 1st minimum is at $0\text{ cm}$. The 6th minimum is at $12.5\text{ cm}$. Number of gaps between the 1st and 6th node $= 6 - 1 = 5$ gaps.

Step 2: Calculate the distance of one gap ($\frac{\lambda}{2}$) $$\text{Distance per gap} = \frac{12.5\text{ cm}}{5} = 2.50\text{ cm}$$ $$\frac{\lambda}{2} = 0.0250\text{ m}$$

Step 3: Calculate the wavelength $\lambda$ $$\lambda = 2 \times 0.0250 = 0.0500\text{ m}$$

Step 4: Calculate the frequency using $v = f\lambda$ $$f = \frac{v}{\lambda} = \frac{3.00 \times 10^8}{0.0500}$$ $$f = 6.00 \times 10^9\text{ Hz} = 6.00\text{ GHz}$$ Answer: $f = 6.00\text{ GHz}$

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Worked example 2 — Speed of Sound in an Air Column

A resonance tube is filled with water, and the water level is lowered while a tuning fork of frequency $480\text{ Hz}$ vibrates over the top. The first loud sound is heard when the air column is $17.2\text{ cm}$ long. The second loud sound is heard when the air column is $52.9\text{ cm}$ long. Calculate the speed of sound in the tube.

Step 1: Identify the relationship between resonance lengths In a closed pipe, the distance between the first resonance ($L_1 = \frac{1}{4}\lambda$) and the second resonance ($L_2 = \frac{3}{4}\lambda$) is exactly half a wavelength. $$L_2 - L_1 = \frac{\lambda}{2}$$

Step 2: Calculate the wavelength $\lambda$ $$\frac{\lambda}{2} = 52.9\text{ cm} - 17.2\text{ cm} = 35.7\text{ cm}$$ $$\lambda = 2 \times 35.7 = 71.4\text{ cm} = 0.714\text{ m}$$

Step 3: Calculate the speed of sound $$v = f\lambda$$ $$v = 480 \times 0.714$$ $$v = 342.72\text{ m s}^{-1}$$

Step 4: Round to significant figures (3 s.f.) Answer: $v = 343\text{ m s}^{-1}$

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Key Equations

Equation	Meaning	Data Sheet?
$v = f\lambda$	Wave speed = frequency $\times$ wavelength	Yes
$d_{N-N} = \frac{\lambda}{2}$	Distance between two adjacent nodes is half a wavelength	No (Memorise)
$d_{N-A} = \frac{\lambda}{4}$	Distance between a node and adjacent antinode is a quarter wavelength	No (Memorise)
$f_n = n f_1$	Frequency of the $n^{\text{th}}$ harmonic (for strings/open pipes)	No (Memorise)
$\Delta \phi = \frac{2\pi x}{\lambda}$	Phase difference between two points on a progressive wave	No (Memorise)

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Common Mistakes to Avoid

• ❌ Wrong: Stating that the distance between two nodes is $\lambda$.
• ✅ Right: The distance between two adjacent nodes is $\frac{1}{2}\lambda$. You must double this distance to find the wavelength.
• ❌ Wrong: Thinking that all points on a stationary wave vibrate with the same amplitude.
• ✅ Right: Amplitude varies from zero at nodes to maximum at antinodes. This is a key difference from progressive waves.
• ❌ Wrong: Claiming that stationary waves transfer energy.
• ✅ Right: Stationary waves store energy in the oscillation. There is no net energy transfer through the medium.
• ❌ Wrong: Assuming points on opposite sides of a node are in phase.
• ✅ Right: All points within one "loop" (between two nodes) are in phase. Points in adjacent loops are $180^\circ$ ($\pi$ rad) out of phase.
• ❌ Wrong: Forgetting to convert centimeters to meters in $v = f\lambda$ calculations.
• ✅ Right: Always use SI units (meters) for wavelength to ensure the speed is in $\text{m s}^{-1}$.

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Exam Tips

1. Describing Formation (The 4-Mark Standard): If an exam question asks how a stationary wave is formed, ensure you include these four points:
   • Two progressive waves travel in opposite directions.
   • The waves have the same frequency (or wavelength).
   • The waves superpose (or overlap/interfere).
   • The resultant displacement is the sum of the individual displacements.
2. Phase Relationships: This is a favorite multiple-choice topic.
   • Points between two nodes: Phase difference = $0$.
   • Points separated by one node: Phase difference = $180^\circ$ ($\pi$ rad).
   • Points separated by two nodes: Phase difference = $0$ (or $360^\circ$).
3. Drawing Stationary Waves: When asked to draw a mode of vibration (e.g., the 2nd harmonic), always draw the two extreme positions of the wave (the "double loop"). Label the nodes (N) and antinodes (A) clearly.
4. Boundary Conditions:
   • Fixed end / Closed end / Metal Reflector: Always a Node.
   • Free end / Open end / Microwave Transmitter: Always an Antinode.
5. The "Difference" Method: In air column experiments, if you are given two successive resonance lengths $L_1$ and $L_2$, always use $(L_2 - L_1) = \frac{\lambda}{2}$. This is more accurate than using $L_1 = \frac{\lambda}{4}$ because it cancels out any "end correction" errors (even though the syllabus says end corrections are negligible, this method is still the standard for high-mark responses).