standing waves sound and acoustic resonance

A standing wave forms when a traveling wave and its reflection from a boundary interfere with the same frequency and opposite propagation directions. The wave equation permits superposition of such waves; for certain system lengths the reflections add in phase and create a stable pattern of nodes and antinodes. Nodes are points where the relevant quantity (displacement or pressure) does not change with time and is zero for that quantity, while antinodes are points of maximum amplitude. Boundary conditions determine the pattern: a fixed string end and a closed pipe end are displacement nodes, whereas an open pipe end is a displacement antinode and a pressure node. Resonance occurs when the driving frequency matches a natural frequency of the system, so energy is efficiently transferred into a high-amplitude standing wave. For a string fixed at both ends, $L=n\frac{\lambda}{2}$ and $f_n=n\frac{v}{2L}$; for a pipe open at both ends, the same formulas apply; for a pipe closed at one end, $L=(2n-1)\frac{\lambda}{4}$ and $f_n=(2n-1)\frac{v}{4L}$, where $n=1,2,3,\ldots$.

A string of length $L=0.80$ m is fixed at both ends. The wave speed on the string is $v=120$ m/s. Find the wavelength and frequency of the fundamental mode, and the frequency of the next mode.

A pipe of length $L=0.25$ m is closed at one end and open at the other. The speed of sound is $v=343$ m/s. Find the first three resonant frequencies, neglecting the end correction.

In a laboratory setup, the sound of a tuning fork produces resonance in a tube with a changing water level. Resonance is heard at air-column lengths $0.17$ m, $0.51$ m, and $0.85$ m. The speed of sound is $343$ m/s. Determine the wavelength and the frequency of the tuning fork. Why are the distances between neighboring resonances equal?

Safe self-study with a simulation.

1. Open an online simulation of standing waves or sound tubes (for example, PhET “Wave Interference” or “Sound”).
2. Choose a string fixed at both ends. By changing the frequency, find the first, second, and third modes; record $L$, $f$, and $v$ from the simulation.
3. Check the formulas $f_n=n\frac{v}{2L}$ and $L=n\frac{\lambda}{2}$.
4. Switch to a pipe closed at one end. Find the first three resonances and check $f_n=(2n-1)\frac{v}{4L}$.
5. Draw a conclusion: which nodes and antinodes are visible at the open and closed ends? Do not use loud sound; when using a real tone, keep the level comfortably low.

Show how boundary conditions select allowed standing-wave modes and where displacement and pressure nodes and antinodes are located.

A diagram with three panels:

• a string fixed at both ends: the fundamental mode and the second mode, with nodes (N) and antinodes (A) marked;
• a pipe open at both ends: pressure nodes and displacement antinodes at the open ends;
• a pipe closed at one end: a displacement node at the closed end and a displacement antinode at the open end.
  Next to the panels, write $L=n\frac{\lambda}{2}$ and $L=(2n-1)\frac{\lambda}{4}$.

A guitar's thickest string is clamped at both ends with a vibrating length $L = 0.65\ \text{m}$ and sounds the note E2 at $f = 82.4\ \text{Hz}$. The standing wave on it has wavelength $\lambda = 2L = 1.30\ \text{m}$, so the wave speed along the string is $v = f\lambda = 82.4\ \text{Hz} \times 1.30\ \text{m} \approx 107\ \text{m/s}$. Pressing at the 12th fret halves the free length to $0.325\ \text{m}$ and the same string sounds $164.8\ \text{Hz}$, one octave higher.

The clamped ends force displacement nodes, so only lengths containing a whole number of half-wavelengths, $L = n\lambda/2$, can persist; $n=1$ is the fundamental and the 12th fret selects $n=1$ of a string half as long. Changing tension changes $v$ through the wave equation and therefore shifts every resonance frequency $f_n = n v/(2L)$.

Blowing across the mouth of a bottle makes the air column inside resonate. A bottle filled with water so that the air column is $L = 0.15\ \text{m}$ long, closed by the water surface and open at the top, sounds near $f = v/(4L) = 343\ \text{m/s}/(4 \times 0.15\ \text{m}) \approx 572\ \text{Hz}$. Pouring in more water shortens the column to $0.10\ \text{m}$ and raises the pitch to about $858\ \text{Hz}$.

One end is a displacement node (water surface) and the other an antinode (open air), so the shortest standing wave fits a quarter wavelength, $L = \lambda/4$, and odd harmonics $f_n = (2n-1)v/(4L)$ follow. The air jet supplies broadband noise, and the column resonates only at frequencies matching its own standing-wave modes.

A car cabin about $L = 2.8\ \text{m}$ long between the windshield and the rear window acts as a tube of air. With $v = 343\ \text{m/s}$, the lowest axial standing wave has $f = v/(2L) = 343\ \text{m/s}/(2 \times 2.8\ \text{m}) \approx 61\ \text{Hz}$. Engine and road noise near $60\ \text{Hz}$ therefore sounds much louder than neighbouring frequencies, and opening a rear window by a few centimetres shifts the resonance and reduces the boom.

Sound waves reflected from the glass superpose with the incoming waves; when the cabin length matches a whole number of half-wavelengths, $L = n\lambda/2$, constructive interference builds large pressure antinodes at some points and nodes at others. This is acoustic resonance of a bounded air volume, the same standing-wave condition as in a pipe.