Frequenze proprie della membrana circolare en
From "Physics, Waves, Music": a website about the physics of waves and sound, acoustics of musical instruments, musical scales, harmony and music.
| AI-translated from Italian. Original page: Frequenze_proprie_della_membrana_circolare |
Definitions and approximations
An ideal circular membrane corresponds to an ideal string, except that its modes of vibration have a different symmetry.
The parameters that describe the membrane are summarised in the following table:
| symbol | meaning | SI unit |
|---|---|---|
| R | radius of the membrane | m |
| T | tension of the membrane | N/m |
| μ | surface density of the membrane | kg m-2 |
Membrane fixed at the edge
The fundamental frequency is given by
The higher frequencies are not in a harmonic series, but have the following ratios to the fundamental. The corresponding modes are, in order, (0,1), (1,1), (2,1), (0,2), (3,1), (1,2), where the first number counts the nodal diameters and the second the nodal circles, edge included (see the page wave equation for membranes).
| partial n | |
|---|---|
| 1 | 1 |
| 2 | 1.59 |
| 3 | 2.14 |
| 4 | 2.30 |
| 5 | 2.65 |
| 6 | 2.92 |
Instruments that use membranes fixed at the edges
The circular shape is the only one used for membranes in musical instruments such as all types of drum, like the snare drum, the timpani (see the section percussion), and the banjo (see the section chordophones).
Real membranes
The formulas above hold for an ideal membrane vibrating in a vacuum. In a real drum the vibrating membrane has to move the surrounding air: the air acts as an additional mass and lowers the frequencies, especially those of the lowest modes. In timpani this effect, together with that of the air enclosed in the kettle, makes the frequencies of the modes (1,1), (2,1), (3,1) and (4,1) stand almost in the ratios 1 : 1.5 : 2 : 2.5, that is like harmonics 2, 3, 4 and 5 of the same series: this is why timpani, unlike most drums, produce a note of well-defined pitch[1].
Further reading and links
- Related questions and answers:
Check your understanding
Problem. The head of a tom-tom has a radius of 15 cm, a tension of 3000 N/m and a surface density of 0.26 kg/m². Calculate the fundamental frequency and that of the second partial.
Solution. = √(3000/0.26) m/s ≈ 107 m/s, so = 0.382 · 107/0.15 Hz ≈ 274 Hz. The second partial is 1.59 times the fundamental, that is about 436 Hz: it is not the octave (548 Hz), as it would be for a string.Problem. A drum has a head with the same tension and the same density as in the previous problem, but twice the radius. What is its fundamental frequency?
Solution. The frequency is inversely proportional to the radius: doubling R halves it, to about 137 Hz, an octave lower. This is why larger drums sound lower.Notes
- ↑ T. D. Rossing, "The Physics of Kettledrums", Scientific American 247(5), 172-178 (1982), doi:10.1038/scientificamerican1182-172
