Frequenze proprie della barra en

AI-translated from Italian. Original page: Frequenze_proprie_della_barra

Definitions and approximations

An ideal bar, in this context, is an object with a uniform cross-section, symmetric with respect to the central plane, straight at rest, and in which, unlike the string, the tension (that is, the longitudinal internal force) is negligible compared with the bending stiffness (that is, the internal forces that oppose bending).

The parameters that describe the bar are summarised in the following table:

symbol	meaning	SI unit
L	length of the bar	m
R	radius of gyration of the bar	m
Y	Young's modulus of the bar	N m^-2
ρ	density of the bar	kg m^-3

The radius of gyration is a property of the cross-section of the bar that describes how the material is distributed with respect to bending: the further the material is from the centre of the cross-section, the stiffer the bar. Here we are interested in rotations of the cross-section about an axis perpendicular to the length of the bar, also perpendicular to the direction of vibration, and passing through the centre of the cross-section. If J is the moment of inertia of the cross-section about this axis and A the area of the cross-section, then $R^{2} = J / A$ . Radii of gyration for various cross-sections are given in the following table

cross-section	radius of gyration R	meaning of a
rectangular	$\frac{a}{\sqrt{12}}$	thickness in the direction of vibration
circular	$\frac{a}{2}$	radius
hollow circular	$\frac{\sqrt{a^{2} + b^{2}}}{2}$	a inner radius, b outer radius

Transverse vibrations

Deriving the natural frequencies of oscillation of a bar is not as easy as for the string, and requires some more mathematics. A complete derivation is given on the page Wave equation in the bar.

Free bar

partial n	$f_{n} / f_{1}$
1	1
2	2.76
3	5.40
4	8.93
5	13.34
6	18.64

The fundamental frequency is given by

ν_{0} = 3.561 \frac{R}{L^{2}} \sqrt{\frac{Y}{ρ}}

The higher frequencies are not in a harmonic series (see the table alongside). You can listen to the frequencies of the bar and see the corresponding normal modes on the page normal modes of a bar (in Italian).

Examples of instruments that use free bars: marimba, xylophone, glockenspiel (see the section percussion)

Bar fixed at one end

partial n	$f_{n} / f_{1}$
1	1
2	6.27
3	17.55
4	34.39
5	56.84
6	84.91

The fundamental frequency is given by

ν_{0} = 0.559 \frac{R}{L^{2}} \sqrt{\frac{Y}{ρ}}

The higher frequencies are not in a harmonic series (see the table alongside). You can listen to the frequencies of the bar and see the corresponding normal modes on the page normal modes of a bar (in Italian).

Examples of instruments that use bars (lamellae) fixed at one end in the form of reeds: harmonica, clarinet, saxophone (see the section aerophones)

Longitudinal vibrations

The fundamental frequency is given by

ν_{0} = \frac{1}{2 L} \sqrt{\frac{Y}{ρ}}

The higher frequencies are in integer ratios with the fundamental, like those of a string.

Check your understanding

Problem. An aluminium bar (Y = 70 GPa, ρ = 2700 kg/m³) 20 cm long, with a rectangular cross-section 1 cm thick in the direction of vibration, is supported so that it vibrates freely, like a glockenspiel bar. Calculate the fundamental frequency and the next two partials.

Solution. The radius of gyration is

R = a / \sqrt{12}

= 0.01/3.464 m ≈ 2.89 mm and

\sqrt{Y / ρ}

≈ 5090 m/s. The fundamental is

ν_{0}

= 3.561 · 0.00289 · 5090/0.2² Hz ≈ 1310 Hz. The next partials are 2.76 and 5.40 times the fundamental, that is about 3610 Hz and 7070 Hz: they are not harmonics.

Problem. The length of the bar of the previous problem is halved. By how much does the note rise?

Solution. The frequency is proportional to

1 / L^{2}

: halving the length makes the frequency four times larger, that is it rises by two octaves (and not by just one, as would happen for a string).

Problem. What is the fundamental frequency of the longitudinal vibrations of the same 20 cm aluminium bar? Compare it with that of the transverse vibrations.

Solution.

ν_{0} = \frac{1}{2 L} \sqrt{Y / ρ}

= 5090/0.4 Hz ≈ 12700 Hz, almost ten times the fundamental of the transverse vibrations (1310 Hz). This is why in struck bars it is mainly the transverse vibrations that are heard.

Physics, Waves, Music

Frequenze proprie della barra en

From "Physics, Waves, Music": a website about the physics of waves and sound, acoustics of musical instruments, musical scales, harmony and music.

Definitions and approximations

Transverse vibrations

Free bar

Bar fixed at one end

Longitudinal vibrations

Further reading and links

Check your understanding

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