AcademySound
Academy
Acoustic Resonance
Level 1 - Physics topic page in Sound.
Principle
Acoustic resonance occurs when driving efficiently excites an allowed air-column mode.
Notation
\(f\)
driving frequency
\(\mathrm{Hz}\)
\(f_n\)
natural frequency
\(\mathrm{Hz}\)
\(L\)
resonator length
\(\mathrm{m}\)
\(v\)
sound speed
\(\mathrm{m\,s^{-1}}\)
\(Q\)
quality factor
1
Method
A resonator stores acoustic energy in a standing wave. The response is largest when the drive matches one of the allowed frequencies.
Resonance condition
\[f=f_n\]
Open-open modes
\[f_n=\frac{nv}{2L}\]
Open-closed modes
\[f_n=\frac{(2n-1)v}{4L}\]
Sharpness measure
\[Q=\frac{f_0}{\Delta f}\]
\Delta f is the frequency width of the strong-response region.
Changing the effective length changes the resonant frequencies. Changing temperature changes them through the sound speed.
Rules
These are the compact resonance relations.
Resonance
\[f=f_n\]
Open-open modes
\[f_n=\frac{nv}{2L}\]
Open-closed modes
\[f_n=\frac{(2n-1)v}{4L}\]
Quality factor
\[Q=\frac{f_0}{\Delta f}\]
Examples
Question
A closed tube resonates at its fundamental when driven at
\[170\,\mathrm{Hz}\]
Estimate its length using \[v=340\,\mathrm{m\,s^{-1}}\]
Answer
For an open-closed fundamental,
\[f_1=\frac{v}{4L}\]
so \[L=\frac{v}{4f_1}=\frac{340}{4(170)}=0.50\,\mathrm{m}\]
Checks
- Resonance amplifies an allowed mode; it does not create arbitrary frequencies.
- Effective length can differ from the physical tube length near open ends.
- A higher \(Q\) means a narrower resonance peak.
- Strong resonance still needs energy input from the driver.