Frequenze proprie dei piatti circolari 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_dei_piatti_circolari |
Definitions and approximations
Plates are the two-dimensional equivalent of the bar: they have their own stiffness, but are not under tension.
The parameters that describe a plate are summarised in the following table.
| symbol | meaning | SI unit |
|---|---|---|
| R | radius of the plate | m |
| s | thickness of the plate | m |
| σ | Poisson's ratio of the plate | dimensionless |
| Y | Young's modulus of the plate | N m-2 |
| ρ | density | kg m-3 |
Plates clamped 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 (as for the circular membrane)[1].
| partial n | |
|---|---|
| 1 | 1 |
| 2 | 2.08 |
| 3 | 3.41 |
| 4 | 3.89 |
| 5 | 5.00 |
| 6 | 5.95 |
Examples of devices that use plates clamped at the edge: telephone receivers (the diaphragm of condenser microphones, instead, is under tension and behaves as a membrane).
Plates fixed at the centre
The fundamental frequency is given by
Examples of instruments that use plates fixed at the centre: cymbals (see the section percussion)
Free plates
The fundamental frequency is given by
The coefficient depends slightly on Poisson's ratio: the value given holds for ≈ 0.3[1]. In the fundamental mode the plate has two perpendicular nodal diameters.
The higher frequencies are not in a harmonic series, and their ratios differ from those of the plate clamped at the edge (for the first modes about 1 : 1.73 : 2.33: two nodal diameters, one nodal circle, three nodal diameters).
Examples of instruments that use free plates: gong (see the section percussion)
Check your understanding
Problem. The steel plate of a telephone receiver is clamped at the edge and has a radius of 2 cm and a thickness of 0.2 mm (Y = 200 GPa, ρ = 7800 kg/m³, σ = 0.3). Calculate the fundamental frequency and that of the second partial.
Solution. = √(2·1011/(7800 · 0.91)) m/s ≈ 5310 m/s and = 0.0002/0.0004 m−1 = 0.5 m−1. So = 0.467 · 0.5 · 5310 Hz ≈ 1240 Hz, in the middle of the voice band transmitted by the telephone. The second partial is 2.08 times the fundamental, that is about 2580 Hz.Problem. How does the fundamental frequency of the plate of the previous problem change if its thickness is doubled? And if its radius is doubled instead?
Solution. The frequency is proportional to . Doubling the thickness doubles the frequency (it rises by an octave), because the plate becomes stiffer. Doubling the radius makes the frequency four times smaller (it drops by two octaves), whereas for a circular membrane at the same tension it would drop by only one octave.Problem. The plate of the first problem is detached from its edge and left free to vibrate. What is its fundamental frequency now?
Solution. Only the coefficient changes: = 1240 · 0.241/0.467 Hz ≈ 640 Hz. The free plate sounds about an octave below the same plate clamped at the edge, because the free edge can move and the plate bends more easily.Further reading and links
- Related questions and answers:
Notes
- ↑ 1,0 1,1 N. H. Fletcher, T. D. Rossing, The Physics of Musical Instruments, 2nd ed., Springer, New York (1998), ch. 3, doi:10.1007/978-0-387-21603-4
