Stationary Waves

When a traveling wave encounters a discontinuity between two different physical media, part of the incident wave energy is reflected back into the first medium, and part is transmitted into the second medium.

1. Boundary Conditions at the Interface

Consider two semi-infinite stretched strings joined seamlessly at $x = 0$ under constant uniform tension $T$. Medium 1 ($x < 0$) has linear density $\mu_1$ and wave speed $v_1 = \sqrt{T/\mu_1}$. Medium 2 ($x > 0$) has linear density $\mu_2$ and wave speed $v_2 = \sqrt{T/\mu_2}$. An incident harmonic wave travels in medium 1: $$y_i(x, t) = A_i \cos(k_1 x - \omega t)$$ Upon striking $x = 0$, it gives rise to a reflected wave $y_r(x, t)$ and a transmitted wave $y_t(x, t)$: $$y_r(x, t) = A_r \cos(-k_1 x - \omega t) = A_r \cos(k_1 x + \omega t)$$ $$y_t(x, t) = A_t \cos(k_2 x - \omega t)$$ Because both sides vibrate at the driving source frequency, angular frequency $\omega$ is identical in both media. The physical interface requires two fundamental boundary conditions at $x = 0$:

1. Displacement Continuity: The string must not tear: $$y_1(0, t) = y_2(0, t) \implies y_i(0, t) + y_r(0, t) = y_t(0, t)$$ $$A_i + A_r = A_t$$
2. Transverse Force Continuity: By Newton's third law, the transverse vertical tension components must balance: $$T \left( \frac{\partial y_1}{\partial x} \right)_{x=0} = T \left( \frac{\partial y_2}{\partial x} \right)_{x=0} \implies k_1 (A_i - A_r) = k_2 A_t$$

2. Amplitude Reflection and Transmission Coefficients

Solving the two linear equations: $$A_r = \left( \frac{k_1 - k_2}{k_1 + k_2} \right) A_i = \left( \frac{v_2 - v_1}{v_1 + v_2} \right) A_i$$ $$A_t = \left( \frac{2 k_1}{k_1 + k_2} \right) A_i = \left( \frac{2 v_2}{v_1 + v_2} \right) A_i$$ Defining the characteristic mechanical wave impedance $Z = \mu v = \sqrt{\mu T} = \frac{T}{v}$: $$r = \frac{A_r}{A_i} = \frac{Z_1 - Z_2}{Z_1 + Z_2}, \quad t = \frac{A_t}{A_i} = \frac{2 Z_1}{Z_1 + Z_2}$$

3. Phase Inversion Analysis

• Denser second medium ($\mu_2 > \mu_1 \implies Z_2 > Z_1$): The reflection coefficient is negative ($r < 0$). Since $\cos(k_1 x + \omega t + \pi) = -\cos(k_1 x + \omega t)$, reflection at an acoustically denser medium induces an instantaneous phase reversal of $\pi$ radians (180°)!
• Rarer second medium ($\mu_2 < \mu_1 \implies Z_2 < Z_1$): $r > 0$. The wave reflects in-phase (zero phase shift).
• Transmission: $t > 0$ always. The transmitted wave never undergoes phase inversion.

4. Conservation of Wave Energy Flux

Wave power is proportional to $Z A^2$: $$P_i = \frac{1}{2} Z_1 \omega^2 A_i^2, \quad P_r = \frac{1}{2} Z_1 \omega^2 A_r^2, \quad P_t = \frac{1}{2} Z_2 \omega^2 A_t^2$$ Defining the energy reflection coefficient $R = P_r / P_i$ and transmission coefficient $T_{trans} = P_t / P_i$: $$R = \left( \frac{Z_1 - Z_2}{Z_1 + Z_2} \right)^2, \quad T_{trans} = \frac{Z_2}{Z_1} \left( \frac{2 Z_1}{Z_1 + Z_2} \right)^2 = \frac{4 Z_1 Z_2}{(Z_1 + Z_2)^2}$$ $$R + T_{trans} = \frac{(Z_1 - Z_2)^2 + 4 Z_1 Z_2}{(Z_1 + Z_2)^2} = \frac{(Z_1 + Z_2)^2}{(Z_1 + Z_2)^2} = 1$$ Energy is strictly conserved across the boundary.

5. Boundary Conditions for No Reflection: Impedance Matching

When $Z_1 = Z_2$: $$R = 0, \quad T_{trans} = 1$$ Zero energy is reflected back; 100% of the wave power passes unhindered into the second medium. This is the foundational principle of impedance matching, critical in ultrasound transducer design (acoustic gel), anti-reflective optical coatings (quarter-wave dielectric layers), and electrical transmission lines.

A stationary (standing) wave is formed when two identical harmonic waves of equal amplitude and frequency propagate in opposite directions through the same medium.

1. Reflection at a Rigid Fixed End ($x = 0$)

At an infinitely rigid termination ($Z_2 \to \infty$), the displacement must vanish identically for all time: $y(0, t) = 0$. The incident wave is $y_i(x, t) = A \cos(kx - \omega t)$. To cancel $y_i$ at $x = 0$, the reflected wave must have $A_r = -A$: $$y_r(x, t) = -A \cos(kx + \omega t)$$ Superposing the incident and reflected waves: $$y(x, t) = y_i(x, t) + y_r(x, t) = A [\cos(kx - \omega t) - \cos(kx + \omega t)]$$ Using the prosthaphaeresis identity $\cos(\alpha - \beta) - \cos(\alpha + \beta) = 2\sin\alpha\sin\beta$: $$y(x, t) = 2 A \sin(kx) \sin(\omega t)$$ Notice the spatial and temporal variables are completely separated! The amplitude of oscillation at any position $x$ is: $$A_{standing}(x) = 2 A |\sin(kx)|$$

2. Nodes and Antinodes

• Nodes: Points of permanent zero displacement ($A_{standing} = 0$): $$\sin(kx) = 0 \implies kx = n\pi \implies x_n = n \frac{\lambda}{2}, \quad n = 0, 1, 2, \dots$$ Consecutive nodes are separated by half a wavelength: $\Delta x_{\text{node}} = \frac{\lambda}{2}$.
• Antinodes: Points of maximum displacement amplitude ($A_{standing} = 2A$): $$|\sin(kx)| = 1 \implies kx = \left(n + \frac{1}{2}\right)\pi \implies x_a = \left(n + \frac{1}{2}\right) \frac{\lambda}{2}$$ Consecutive antinodes are separated by $\frac{\lambda}{2}$. The distance between an adjacent node and antinode is a quarter wavelength: $\frac{\lambda}{4}$.

3. Reflection at a Free End ($x = 0$)

At an unconstrained frictionless ring / free end, transverse force vanishes: $\frac{\partial y}{\partial x}\Big|_{x=0} = 0$. Here $A_r = +A$ (no phase flip): $$y(x, t) = A [\cos(kx - \omega t) + \cos(kx + \omega t)] = 2 A \cos(kx) \cos(\omega t)$$ An antinode forms directly at the free boundary.

When a string of length $L$ is clamped rigidly at both ends ($x = 0$ and $x = L$), boundary conditions permit only discrete resonant modes of oscillation, known as normal modes.

1. Boundary Conditions and Mode Frequencies

The general standing wave solution is $y(x, t) = [C_1 \sin(kx) + C_2 \cos(kx)] \cos(\omega t + \phi)$.

1. At $x = 0$: $y(0, t) = 0 \implies C_2 = 0$. Thus $y(x, t) = C_1 \sin(kx) \cos(\omega t + \phi)$.
2. At $x = L$: $y(L, t) = 0 \implies C_1 \sin(kL) = 0$.

For non-trivial oscillations ($C_1 \ne 0$): $$\sin(k L) = 0 \implies k_n L = n \pi, \quad n = 1, 2, 3, \dots$$ The allowed wavenumbers and wavelengths are: $$k_n = \frac{n\pi}{L}, \quad \lambda_n = \frac{2L}{n}$$ Since wave speed is $v = \sqrt{T/\mu}$, the allowed proper (eigen) frequencies are: $$f_n = \frac{v}{\lambda_n} = \frac{n v}{2L} = \frac{n}{2L}\sqrt{\frac{T}{\mu}}, \quad n = 1, 2, 3, \dots$$

2. Harmonic Spectrum of a Stretched String

• Fundamental Mode / First Harmonic ($n = 1$): $$\lambda_1 = 2L, \quad f_1 = \frac{1}{2L}\sqrt{\frac{T}{\mu}}$$ Contains 2 nodes at the ends and 1 central antinode.
• Second Harmonic / First Overtone ($n = 2$): $$\lambda_2 = L, \quad f_2 = 2 f_1$$ Contains 3 nodes (including center $x = L/2$) and 2 antinodes.
• Third Harmonic / Second Overtone ($n = 3$): $$\lambda_3 = \frac{2L}{3}, \quad f_3 = 3 f_1$$

Because all integer multiples $f_n = n f_1$ are present, the overtone spectrum is complete and richly consonant, giving musical string instruments (violin, guitar, piano) their warm, harmonic timbre.

3. Mersenne's Laws of Vibrating Strings

Marin Mersenne (1636) summarized the physical dependencies of the fundamental frequency:

1. Law of Length: $f \propto \frac{1}{L}$ (halving length doubles pitch).
2. Law of Tension: $f \propto \sqrt{T}$ (quadrupling tension doubles pitch).
3. Law of Density: $f \propto \frac{1}{\sqrt{\mu}} = \frac{1}{r \sqrt{\rho}}$ (thicker strings produce lower pitch).

In acoustic columns (organ pipes, flutes, clarinets), standing waves are formed by longitudinal air particle displacements and pressure oscillations.

1. Displacement vs. Pressure Waves

In a sound wave, particle displacement $s(x, t)$ and acoustic gauge pressure $p(x, t)$ are out of phase by 90°: $$s(x, t) = s_0 \sin(kx - \omega t) \implies p(x, t) = -B \frac{\partial s}{\partial x} = -B k s_0 \cos(kx - \omega t)$$ Consequently:

• A displacement node (rigid closed boundary where air cannot move) is always a pressure antinode (maximum pressure variation).
• A displacement antinode (open pipe end open to atmosphere) is always a pressure node (pressure is fixed at atmospheric $p = 0$).

2. Open Pipe (Open at Both Ends)

Both ends are open to atmosphere $\implies$ displacement antinodes at both $x = 0$ and $x = L$: $$\lambda_n = \frac{2L}{n}, \quad f_n = \frac{n v}{2L} = n f_1, \quad n = 1, 2, 3, \dots$$ An open pipe produces all harmonics (both even and odd).

3. Closed Pipe (Closed at One End, Open at the Other)

Closed end at $x = 0$ (displacement node); open end at $x = L$ (displacement antinode): $$L = (2n - 1) \frac{\lambda_n}{4} \implies \lambda_n = \frac{4L}{2n - 1}$$ $$f_n = \frac{(2n - 1) v}{4L} = (2n - 1) f_1, \quad n = 1, 2, 3, \dots$$ A closed pipe produces only odd harmonics ($f_1, 3f_1, 5f_1, \dots$). The fundamental frequency of a closed pipe is half that of an open pipe of identical length ($f_{1,\text{closed}} = \frac{1}{2} f_{1,\text{open}}$).

4. End Correction

In reality, the acoustic wave reflects slightly outside the open end of a tube of radius $R$. Lord Rayleigh showed that an acoustic end correction $e \approx 0.61 R$ must be added for each open end:

• Open Pipe: $L_{eff} = L + 2(0.61 R) = L + 1.22 R$.
• Closed Pipe: $L_{eff} = L + 0.61 R$.

Franz Melde (1859) devised an elegant electro-mechanical apparatus to demonstrate standing waves and verify Mersenne's laws.

1. Experimental Arrangements of Melde's Apparatus

A light string of length $L$ and linear density $\mu$ is tied to one prong of an electrically driven tuning fork of frequency $f_{fork}$ and stretched horizontally over a frictionless pulley with suspended mass $M$ ($T = M g$).

1. Transverse Arrangement: The prongs vibrate perpendicular to the length of the string. For every oscillation of the fork prong, the string is displaced once: $$f_{\text{string}} = f_{fork}$$ If the string vibrates in $p$ resonant loops: $L = p \frac{\lambda}{2} \implies \lambda = \frac{2L}{p}$. $$f_{fork} = \frac{p}{2L}\sqrt{\frac{T}{\mu}} = \frac{p}{2L}\sqrt{\frac{M g}{\mu}} \implies \frac{T}{p^2} = \text{constant}$$
2. Longitudinal Arrangement: The prongs vibrate parallel to the length of the string. When the prong moves forward, tension drops; when it moves backward, tension peaks. The string is pulled twice during each complete cycle of the tuning fork. Hence the frequency of the string is exactly half the tuning fork frequency (parametric excitation): $$f_{\text{string}} = \frac{1}{2} f_{fork}$$ $$f_{fork} = \frac{p}{L}\sqrt{\frac{T}{\mu}}$$

2. Energy Distribution in Stationary Waves

In a traveling wave, energy flows continuously downstream. In a stationary wave: $$y(x, t) = 2 A \sin(kx) \cos(\omega t)$$

• Kinetic Energy Density: $$u_K(x, t) = \frac{1}{2}\mu \left(\frac{\partial y}{\partial t}\right)^2 = 2 \mu \omega^2 A^2 \sin^2(kx) \sin^2(\omega t)$$
• Potential Energy Density: $$u_P(x, t) = \frac{1}{2}T \left(\frac{\partial y}{\partial x}\right)^2 = 2 T k^2 A^2 \cos^2(kx) \cos^2(\omega t) = 2 \mu \omega^2 A^2 \cos^2(kx) \cos^2(\omega t)$$

Notice:

• When $\sin(\omega t) = 1$ (string passing through equilibrium $y = 0$), potential energy is zero everywhere, and total energy resides purely as kinetic energy concentrated at the antinodes ($\sin(kx) = 1$).
• When $\cos(\omega t) = 1$ (string at maximum displacement), kinetic energy is zero everywhere, and total energy resides purely as potential elastic energy concentrated at the nodes ($\cos(kx) = 1$, where string slope is steepest!).

Total energy remains permanently trapped, surging periodically between nodes (elastic potential energy) and antinodes (kinetic energy) with zero net spatial flux across the nodes.