Topic 9.6.114 min read
Stationary Waves in a Stretched String
A string fixed at both ends supports only those stationary-wave patterns for which its length contains a whole number of half-wavelengths. These allowed modes have discrete harmonic frequencies that are whole-number multiples of the fundamental frequency.
Choose Your Level
Beginner Level
A string tied at both ends cannot vibrate in just any shape. Both ends must stay still, so they are nodes. The simplest pattern has one loop; faster patterns have two, three, or more loops.
Recommended for Class 11
Overview
A string fixed at both ends supports only those stationary-wave patterns for which its length contains a whole number of half-wavelengths. These allowed modes have discrete harmonic frequencies that are whole-number multiples of the fundamental frequency.
Learning Outcomes
- Both fixed ends of the stretched string are nodes.
- One loop is formed in the fundamental mode, two loops in the second harmonic, and three loops in the third harmonic.
- For n loops, the string length contains n half-wavelengths: L=nλ_n/2.
- The wavelength of the nth mode is λn=2L/n.
- The frequency of the nth mode is fn=nv/(2L)=nf1.
- For a string under tension T with mass per unit length m, v=√(Tm).
- The fundamental frequency is the lowest allowed frequency; its integer multiples are harmonics or overtones.
- The discrete set of allowed frequencies illustrates quantization of frequency.
Intermediate Level
String Fixed at Both Ends
Consider a string of length L stretched by tension T and fixed at both ends. A transverse wave reflects at each fixed end and superposes with the incoming wave. Because a fixed end cannot move, both ends must be nodes. Only patterns that place nodes at both ends can persist as stationary waves.
Fundamental Mode: One Loop
When the string is plucked at its middle, the simplest stationary pattern contains one loop. The ends are nodes and the middle is an antinode. One loop is one half-wavelength, so the string length equals half of λ₁.
This lowest allowed frequency f₁ is called the fundamental frequency, first harmonic, or first mode of vibration.
Wave Speed and Fundamental Frequency
If M is the total mass of the string, then its mass per unit length is m=M/L. The transverse-wave speed depends on the tension and linear mass density.
| Symbol | Meaning | SI unit |
|---|---|---|
| L | Length of string | m |
| T | Tension in string | N |
| m | Mass per unit length, M/L | kg m⁻¹ |
| v | Wave speed on string | m s⁻¹ |
| n | Mode or harmonic number | no unit |
Second Harmonic: Two Loops
When the string vibrates in two loops, each loop still occupies half a wavelength. The complete wavelength λ₂ fits into the string, and the frequency is twice the fundamental frequency.
Third Harmonic: Three Loops
For three loops, the string contains three half-wavelengths. Its wavelength is 2L/3 and its frequency is three times the fundamental.
General nth Harmonic
If the string vibrates in n loops, n half-wavelengths fit exactly into its length. This gives the general resonance, wavelength and frequency relations.
| Mode | Loops | Wavelength | Frequency | Name |
|---|---|---|---|---|
| n=1 | 1 | 2L | f₁ | Fundamental / first harmonic |
| n=2 | 2 | L | 2f₁ | Second harmonic / first overtone |
| n=3 | 3 | 2L/3 | 3f₁ | Third harmonic / second overtone |
| n | n | 2L/n | nf₁ | nth harmonic |
Resonance and Quantization
The string resonates only when its length is a whole-number multiple of half a wavelength. Therefore, it cannot sustain every possible frequency; it supports only the discrete set f₁, 2f₁, 3f₁, and so on. This discrete selection is called quantization of frequency.
How the Mode Number Changes the Wave
| As n increases | Result |
|---|---|
| Number of loops | Increases |
| Frequency fₙ | Increases |
| Wavelength λₙ | Decreases |
| Wave speed v=fₙλₙ | Remains constant for unchanged T and m |
Third Harmonic of a String
A 1.20 m string carries waves at 240 m s⁻¹. Find the wavelength and frequency of its third harmonic.
- L=1.20m
- v=240m s-1
- n=3
λ3=2L3=2(1.20)3=0.80m
f3=3v2L=3(240)2(1.20)=300Hz
Final answer: The third-harmonic wavelength is 0.80 m and its frequency is 300 Hz.
Common Mistakes
- Both fixed ends are nodes, not antinodes.
- One loop equals λ/2, not one full wavelength.
- The first overtone is the second harmonic, not the first harmonic.
- For a fixed string, fₙ increases with n while λₙ decreases.
- Use mass per unit length in v=√(Tm), not the total mass by itself.
- The symbol T in this section means tension, not time period.
Short-Question Preparation
Q. What is the fundamental frequency of a stretched string?
A. It is the lowest allowed frequency, f₁=v2L, produced when the string vibrates in one loop.
Q. What condition must a fixed string satisfy for resonance?
A. Its length must contain a whole number of half-wavelengths: L=nλₙ/2.
Q. What is frequency quantization on a string?
A. Only the discrete frequencies f₁, 2f₁, 3f₁, … are allowed.
Q. How are harmonics related to the fundamental?
A. The nth harmonic has frequency fₙ=nf₁.
Sticky Notes
- Both fixed ends of the stretched string are nodes.
- One loop is formed in the fundamental mode, two loops in the second harmonic, and three loops in the third harmonic.
- For n loops, the string length contains n half-wavelengths: L=nλ_n/2.
- The wavelength of the nth mode is λn=2L/n.
- The frequency of the nth mode is fn=nv/(2L)=nf1.
- For a string under tension T with mass per unit length m, v=√(Tm).
- The fundamental frequency is the lowest allowed frequency; its integer multiples are harmonics or overtones.
- The discrete set of allowed frequencies illustrates quantization of frequency.
60-Second Revision
A string fixed at both ends has nodes at both ends. With n loops, L=nλₙ/2, so λₙ=2L/n and fₙ=nv/(2L)=nf₁. For wave speed v=√(Tm), fₙ=n2L√(Tm). The allowed frequencies are discrete harmonics; higher n means more loops, shorter wavelength and higher frequency.
Summary
A string fixed at both ends supports only those stationary-wave patterns for which its length contains a whole number of half-wavelengths. These allowed modes have discrete harmonic frequencies that are whole-number multiples of the fundamental frequency.
Ask-the-Tutor Modes
- Draw and explain the first three modes.
- Derive λₙ=2L/n.
- Derive fₙ=nv/(2L).
- Explain harmonics and overtones.
- Solve a stretched-string numerical.
Exam Focus
Exam Focus
Important points
- Both fixed ends are nodes.
- L=nλₙ/2 and λₙ=2L/n.
- fₙ=nv/(2L)=nf₁.
- v=√(Tm).
- Fundamental, harmonics, overtones and quantization.
Important formulas
- L=nλn2
- λn=2Ln
- fn=nv2L=nf1
- v=√Tm
- fn=n2L√Tm
Common exam traps
- One loop is half a wavelength.
- First overtone means second harmonic.
Long question structure
Fixed-end condition → fundamental → second and third modes → nth harmonic → quantization and overtones
Common Mistakes
Common Mistakes
- Mistake
Both fixed ends are nodes.
- Mistake
One loop equals λ/2.
- Mistake
The first overtone is the second harmonic.
- Mistake
Use linear mass density in the string-speed formula.
- Mistake
T denotes tension in these equations.
Revision Zone
60-Second Revision
Quick cues from this topic — use them for last-minute revision.
Both fixed ends of the stretched string are nodes.
One loop is formed in the fundamental mode, two loops in the second harmonic, and three loops in the third harmonic.
For n loops, the string length contains n half-wavelengths: L=nλ_n/2.
The wavelength of the nth mode is λn=2L/n.
A string fixed at both ends supports only those stationary-wave patterns for which its length contains a whole number of half-wavelengths. These allowed modes have discrete harmonic frequencies that are whole-number multiples of the fundamental frequency.
- Definition
- A string fixed at both ends supports only those stationary-wave patterns for which its length contains a whole number of half-wavelengths. These allowed modes have discrete harmonic frequencies that are whole-number multiples of the fundamental frequency.
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Exam Practice
Important Question
Q1. Derive the wavelength and frequency of the fundamental mode of a string fixed at both ends.
Important Question
Q2. Derive the general frequency of the nth harmonic.
Important Question
Q3. Explain frequency quantization for a stretched string.
Important Question
Q4. Distinguish between harmonics and overtones.
Important Question
Q5. State how length, tension and linear density affect string frequency.