C4-2. Harmonics in Strings and Pipes

1. Harmonics on a Stretched String

Boundary Conditions: Fixed at Both Ends

When a progressive transverse wave travels along a string fixed at both ends, it undergoes reflection at the boundaries. Because the ends are clamped tightly, the reflected wave experiences an instantaneous phase inversion of $\pi$ radians ($180^\circ$) relative to the incident wave. This phase inversion leads to continuous, complete destructive interference at the fixed walls, forcing a permanent zero-displacement point—or node (N)—to form at each boundary.

  • Because the ends are fixed, a node must form at both ends.
  • The length $L$ must be an integer multiple of a half-wavelength.
  • All integer harmonics exist ($n = 1, 2, 3, \dots$).
  • HL Concept: The speed of the progressive waves forming the standing wave depends on the mechanical tension ($T$) and the linear mass density ($\mu$): $v = \sqrt{\dfrac{T}{\mu}}$.

Notation Key:

  • $n$ = The integer harmonic number ($n = 1$ for fundamental, $n = 2$ for 2nd harmonic, etc.)
  • $\lambda_n$ = Wavelength of the $n$-th harmonic ($\text{m}$)
  • $f_n$ = Frequency of the $n$-th harmonic ($\text{Hz}$)
  • $L$ = Total vibrating length of the string ($\text{m}$)
  • $v$ = Speed of the progressive wave along the string ($\text{m s}^{-1}$)
  • $T$ = Mechanical tension applied to the string ($\text{N}$)
  • $\mu$ = Linear mass density of the string ($\text{kg m}^{-1}$)
$$\lambda_n = \dfrac{2L}{n}$$ $$f_n = \dfrac{nv}{2L} = \dfrac{n}{2L}\sqrt{\dfrac{T}{\mu}}$$
1st Harmonic (Fundamental) λ₁ = 2L f₁ = v / 2L N A N 2nd Harmonic (1st Overtone) λ₂ = L f₂ = 2f₁ = v / L N A N A N 3rd Harmonic (2nd Overtone) λ₃ = 2L / 3 f₃ = 3f₁ = 3v / 2L N A N A N A N Total Length (L)

2. Harmonics in Pipes (Open at Both Ends)

Boundary Conditions: Two Open Ends

  • Because the air is free to vibrate at the openings, an antinode (maximum displacement) must form at both ends.
  • The wave shapes are the inverse of the string, but the mathematics are identical.
  • All integer harmonics exist ($n = 1, 2, 3, \dots$).

Notation Key:

  • $n$ = The integer harmonic number ($n = 1$ for fundamental, $n = 2$ for 1st overtone, etc.)
  • $\lambda_n$ = Wavelength of the $n$-th harmonic ($\text{m}$)
  • $f_n$ = Frequency of the $n$-th harmonic ($\text{Hz}$)
  • $L$ = Total vibrating length of the string ($\text{m}$)
  • $v$ = Speed of the progressive wave along the string ($\text{m s}^{-1}$)
  • $T$ = Mechanical tension applied to the string ($\text{N}$)
  • $\mu$ = Linear mass density of the string ($\text{kg m}^{-1}$)
$$\lambda_n = \dfrac{2L}{n}$$ $$f_n = \dfrac{nv}{2L} = n f_1$$
1st Harmonic (Fundamental) n = 1, λ₁ = 2L f₁ = v / 2L A N A 2nd Harmonic (1st Overtone) n = 2, λ₂ = L f₂ = 2f₁ = v / L A N A N A 3rd Harmonic (2nd Overtone) n = 3, λ₃ = 2L / 3 f₃ = 3f₁ = 3v / 2L A N A N A N A

3. Harmonics in Pipes (Closed at One End)

Boundary Conditions: One Closed, One Open

When a sound wave travels through a column of air closed at one end, the rigid physical wall and the open atmospheric boundary dictate how the standing wave must form. This creates an asymmetric boundary system.

  • A node forms at the closed wall because the air molecules are physically blocked and cannot displace longitudinally.
  • An antinode forms at the open end because the air molecules are completely free to vibrate into the surrounding atmosphere.
  • The fundamental frequency (first harmonic) only fits exactly a quarter wavelength ($\lambda/4$) inside the pipe.
  • Key Trap: Because the boundaries are mismatched (Node to Antinode), it is mathematically impossible to fit an even number of quarter-wavelengths into the tube. Therefore, only odd harmonics exist ($n = 1, 3, 5, \dots$). There is no 2nd, 4th, or 6th harmonic in a closed pipe.
$$\lambda_n = \dfrac{4L}{n}$$ $$f_n = \dfrac{nv}{4L} = n f_1 \quad \text{for } n = 1, 3, 5, \dots$$
EXAMINER CORE INSIGHT - Displacement vs. Pressure Waves

In IB HL Physics, you must distinguish between displacement standing waves and pressure standing waves. They are exactly $\dfrac{\pi}{2}$ radians ($90^\circ$) out of phase spatially.

• At the closed end: Air cannot move (Displacement Node), meaning molecules crash into the wall and bunch up, creating maximum pressure variation (Pressure Antinode).
• At the open end: Air moves freely (Displacement Antinode), but because it meets the vast atmosphere, the pressure is locked at a constant 1 atm (Pressure Node).

1st Harmonic (Fundamental) n = 1, λ₁ = 4L f₁ = v / 4L N A 3rd Harmonic (1st Overtone) n = 3, λ₃ = 4L / 3 f₃ = 3f₁ = 3v / 4L N A N A 5th Harmonic (2nd Overtone) n = 5, λ₅ = 4L / 5 f₅ = 5f₁ = 5v / 4L N A N A N A

Example 1: Buoyancy and Mechanical Tension Revisions

Problem: A steel wire of length $L = 0.80\text{ m}$ is clamped at one end. The other end passes over a frictionless pulley and is attached to a solid metal block of mass $M = 2.50\text{ kg}$ and volume $V = 4.00 \times 10^{-4}\text{ m}^3$ hanging freely in the air. In this arrangement, the wire vibrates in its fundamental mode at a frequency of $220\text{ Hz}$.

A beaker of water (density $\rho_w = 1000\text{ kg m}^{-3}$) is then raised so that the metal block is completely submerged. Calculate the new fundamental frequency of the vibrating wire. (Assume $g = 9.81\text{ m s}^{-2}$).

Solution:

Step 1: Understand the Functional Proportionality
The fundamental frequency $f_1$ of a fixed-fixed string is given by $f_1 = \dfrac{1}{2L}\sqrt{\dfrac{T}{\mu}}$. Since the physical length $L$ and the linear mass density $\mu$ remain perfectly static throughout this scenario, the frequency depends exclusively on the square root of the active tension ($f_1 \propto \sqrt{T}$). Therefore, we can set up a direct ratio:

$$\dfrac{f_{\text{new}}}{f_{\text{old}}} = \sqrt{\dfrac{T_{\text{new}}}{T_{\text{old}}}}$$

Step 2: Compute Initial Mechanical Tension in Air ($T_{\text{old}}$)
While hanging freely in the atmosphere, the upward pulling tension balances the downward weight of the block exactly:

$$T_{\text{old}} = Mg = (2.50\text{ kg})(9.81\text{ m s}^{-2}) = 24.525\text{ N}$$

Step 3: Compute Submerged Tension incorporating Upward Buoyancy ($T_{\text{new}}$)
When completely immersed, Archimedes' principle dictates that an upward buoyant force ($F_B$) acts on the block, equal to the weight of the water volume displaced:

$$F_B = \rho_w V g = (1000\text{ kg m}^{-3})(4.00 \times 10^{-4}\text{ m}^3)(9.81\text{ m s}^{-2}) = 3.924\text{ N}$$

Using the equilibrium condition for the submerged block ($T_{\text{new}} + F_B = Mg$):

$$T_{\text{new}} = Mg - F_B = 24.525\text{ N} - 3.924\text{ N} = 20.601\text{ N}$$

Step 4: Determine the Final Modulated Frequency

$$f_{\text{new}} = f_{\text{old}} \times \sqrt{\dfrac{T_{\text{new}}}{T_{\text{old}}}} = 220\text{ Hz} \times \sqrt{\dfrac{20.601}{24.525}} = 220 \times \sqrt{0.8400} = \mathbf{201.6\text{ Hz}}$$

Example 2: Compound Strings and Boundary Interface Nodes

Problem: A heavy copper wire of length $L_1 = 0.60\text{ m}$ and linear mass density $\mu_1 = 8.00 \times 10^{-3}\text{ kg m}^{-1}$ is spliced end-to-end to a lighter aluminum wire of length $L_2 = 0.40\text{ m}$ and linear mass density $\mu_2 = 2.00 \times 10^{-3}\text{ kg m}^{-1}$. The combined string is anchored between two fixed rigid walls under a continuous shared mechanical tension of $T = 180\text{ N}$. Find the lowest frequency (greater than zero) at which a standing wave pattern can form such that the joint linking the two segments forms a node.

Solution:

Step 1: Determine Progressive Wave Speed within Each Material segment
Because the linear mass densities differ, the progressive components propagate at vastly different velocities across the boundary interface:

$$v_1 = \sqrt{\dfrac{T}{\mu_1}} = \sqrt{\dfrac{180\text{ N}}{8.00 \times 10^{-3}\text{ kg m}^{-1}}} = \sqrt{22500} = 150\text{ m s}^{-1}$$ $$v_2 = \sqrt{\dfrac{T}{\mu_2}} = \sqrt{\dfrac{180\text{ N}}{2.00 \times 10^{-3}\text{ kg m}^{-1}}} = \sqrt{90000} = 300\text{ m s}^{-1}$$

Step 2: Impose Boundary Frameworks for Each Section
Since both wall boundaries are nodes, and the connection joint is forced to remain a node, each section must satisfy the standard standing wave criteria independently. Let $n_1$ and $n_2$ symbolize the respective integer harmonic states. The driving frequency $f$ must remain uniform throughout the unified compound system:

$$f = \dfrac{n_1 v_1}{2 L_1} = \dfrac{n_2 v_2}{2 L_2}$$

Step 3: Evaluate the Harmonic Multiplier Ratio
Isolate the relationship between the two integer harmonic indices by introducing our speeds and lengths:

$$\dfrac{n_1 (150)}{2(0.60)} = \dfrac{n_2 (300)}{2(0.40)} \implies 125 n_1 = 375 n_2$$ $$\dfrac{n_1}{n_2} = \dfrac{375}{125} = \dfrac{3}{1}$$

Step 4: Solve for the Lowest Feasible Frequency State
To find the absolute lowest physical frequency ($f > 0$), we take the lowest possible integers matching this exact spatial ratio: $n_1 = 3$ and $n_2 = 1$. Substituting these back yields:

$$f = 125 \times 3 = 375 \times 1 = \mathbf{375\text{ Hz}}$$

Physics Context: At $375\text{ Hz}$, the heavier copper wire forms exactly 3 standing loops (3rd harmonic), while the lighter aluminum segment resonates simultaneously in a single fundamental loop (1st harmonic), matching perfectly with zero amplitude at the joint interface.

Example 3: End Corrections in Pipes (HL)

Problem: In a rigorous resonance tube experiment, a tube open at one end and closed at the other is made to resonate by a tuning fork of frequency $f = 440\text{ Hz}$. Because the air displacement does not stop abruptly at the rigid boundary of the pipe, the displacement antinode actually forms a small distance $c$ (the end correction) outside the physical end of the pipe.

The first position of resonance occurs when the physical length of the air column is $L_1 = 0.180\text{ m}$. As water is drained, the next position of resonance occurs when the length is extended to $L_2 = 0.565\text{ m}$.
Calculate:
(a) The speed of sound in the air column.
(b) The precise value of the end correction, $c$.
(c) Given that the end correction $c$ is approximated by $c \approx 0.6r$, where $r$ is the internal radius of the pipe, calculate the internal diameter of the resonance tube.

Solution:

(a) Speed of sound
The true effective length of the resonating air column is $L_{\text{eff}} = L_{\text{physical}} + c$.
For the first resonance ($n=1$): $\dfrac{\lambda}{4} = L_1 + c \quad$ (Eq. 1)
For the second resonance ($n=3$): $\dfrac{3\lambda}{4} = L_2 + c \quad$ (Eq. 2)
Subtracting Eq. 1 from Eq. 2 entirely eliminates the unknown end correction $c$:

$$\dfrac{3\lambda}{4} - \dfrac{\lambda}{4} = (L_2 + c) - (L_1 + c)$$ $$\dfrac{2\lambda}{4} = \dfrac{\lambda}{2} = L_2 - L_1 = 0.565 - 0.180 = 0.385\text{ m}$$ $$\lambda = 2 \times 0.385 = 0.770\text{ m}$$

The speed of sound is $v = f\lambda = 440\text{ Hz} \times 0.770\text{ m} = \mathbf{338.8\text{ m s}^{-1}}$.

(b) The End Correction ($c$)
Substitute $\lambda = 0.770\text{ m}$ back into the fundamental resonance equation (Eq. 1):

$$\dfrac{0.770}{4} = 0.180 + c$$ $$0.1925 = 0.180 + c \implies c = 0.1925 - 0.180 = \mathbf{0.0125\text{ m} \text{ (or } 1.25\text{ cm})}$$

(c) Internal Diameter Calculation
Using the theoretical approximation linking the end correction to the pipe's internal radius:

$$c = 0.6r \implies 0.0125\text{ m} = 0.6r$$ $$r = \dfrac{0.0125}{0.6} = 0.02083\text{ m}$$

The diameter $d$ is simply twice the radius ($d = 2r$):
$d = 2 \times 0.02083 = 0.04166\text{ m}$. Rounded to 3 significant figures, the diameter of the tube is $4.17\text{ cm}$.