Muusikainstrumentide füüsikalisi aspekte · 2007. 4. 11. · Pingul keele seadus (law of...

Muusikainstrumentide füüsikalisi aspekte

Evely Leetma

Seminariettekanne aines Heli ja muusika matemaatilised aspektid

29. märts 2007

Muusikainstrumentide klassifikatsioon

Keelpillid ehk kordofonidpoogenkeelpillid: viiul, vioola, tšello, kontrabassnäppekeelpillid: harf, kitarr, mandoliin, balalaika, kannellöökkeelpillid: klaver

Puhkpillid ehk aerofonidpuupuhkpillid: flööt, oboe, klarnet, saksofon, fagottvaskpuhkpillid: trompet, metsasarv, tromboon, tuubalõõtspuhkpillid: akordion, harmoonium, orel

Evely Leetma Muusikainstrumentide füüsikalisi aspekte 2 / 25

Muusikainstrumentide klassifikatsioon

LöökpillidMembranofonid

Kindla helikõrgusega: timpanidEbamäärase helikõrgusega: trummid, bongod, tamburiin

IdiofonidKindla helikõrgusega: kellamäng, vibrafon, ksülofon,marimbafonEbamäärase helikõrgusega: triangel, taldrikud,kastanjetid

Elektrofonid: süntesaator


Keele vaba võnkumise võrrand

Olgu keele tasakaaluasendiks x-telg ja keele pikkus l.Tähistagu y(x, t) keele punkti x kõrvalekallettasakaaluasendist ajamomendil t. Olgu T [N = kg m/s2] pingekeele suvalises punktis ja ρ [kg/m] keele lineaarne tihedus.Vaatleme väikesi ristvõnkumisi.

3.2. THE WAVE EQUATION FOR STRINGS 85

3.2. The wave equation for strings

Italian Theorbo, Musée Instrumental, Brussels, Belgium

In this section, we return to the subject of §1.6, and consider the rel-evance of Fourier series to the vibration of a string held at both ends. Tomake a more accurate analysis, we need to regard the displacement y as afunction both of time t and position x along the string. Since y is being re-garded as a function of two variables, the appropriate equations are writtenin terms of partial derivatives, and Appendix P gives a brief summary of par-tial derivatives. The equation describing the vibration of a string is called thewave equation in one dimension, which we now develop. This equation sup-poses that the displacement of the string is such that its slope at any pointalong its length at any time is small. For large displacements, the analysis isharder. Note that we are only concerned here with transverse waves, namelymotion perpendicular to the string. Motion parallel to the string is calledlongitudinal waves, and will be ignored here.

θ(x)

θ(x+∆x)

T

T sin θ(x)

TT sin θ(x+∆x)

Write T for the tension on the string (in newtons = kgm/s2), andρ for the linear density of the string (in kg/m). Then at position x alongthe string, the angle θ(x) between the string and the horizontal will satisfy

tan θ(x) =∂y

∂x. On a small segment of string from x to x+ ∆x, the vertical

component of force at the left end will be −T sin θ(x), and at the right endwill be T sin θ(x+ ∆x).

Provided that θ(x) is small, sin θ(x) and tan θ(x) are approximatelyequal. So the difference in vertical components of force between the two ends



Kasutades fakti, et sin θ ≈ tan θ, kui θ → 0, saame jõuvertikaalsete komponentide erinevuseks

T tan θ(x+ ∆x)− T tan θ(x) ≈ T∆x∂2y

∂x2

Newtoni II seaduse F = ma põhjal, kus m ≈ ρ∆x ja a = ∂2y∂t2

,saame keele võnkumise võrrandi

∂2y

∂t2= c2

∂2y

∂x2,

kus c =√T/ρ.



D’Alembert teisendas võrrandi muutujate vahetuse u = x+ ct,v = x− ct abil kujule

∂2y

∂u∂v= 0

ning otsis üldlahendit kujul

y = f(x+ ct) + g(x− ct).



Rajatingimuste y(0, t) = 0 ja y(l, t) = 0 korral saamelainevõrrandi üldlahendiks

y = f(x+ ct)− f(ct− x),

kus f(λ) = f(λ+ 2l) ∀λ.

http://home.austin.rr.com/jmjensen/JeffString.html

http://id.mind.net/~zona/mstm/physics/waves/standingWaves/standingWaves1/StandingWaves1.html


http://home.austin.rr.com/jmjensen/JeffString.htmlhttp://id.mind.net/~zona/mstm/physics/waves/standingWaves/standingWaves1/StandingWaves1.htmlhttp://id.mind.net/~zona/mstm/physics/waves/standingWaves/standingWaves1/StandingWaves1.html

Pingul keele seadus (law of stretched string)

Marin Mersenne "law of stretched string":

n-nda harmooniku sagedus avaldub kujul

ν =n

2l

√T

ρ,

kus l on keele pikkus, T pinge ja ρ keele lineaarne tihedus.


Ülesanne

Mitu korda tuleb keele pinget suurendada, et saada kvindivõrra kõrgem heli?

Näpunäide:

ν =n

2l

√T

ρ

Puhtale kvindile vastab sageduste suhe 3:2.


Ülesanne

Klaveri keeled on valmistatud terasest tihedusega 5900 kg/m3.Tootjad soovitavad rakendada pinget ligikaudu 1.1 · 109 N/m2.Milline on helilaine levimise kiirus selles keeles? Kas see sõltubristlõike pindalast?

Vaatame kahte keelt ristlõike pindaladega 1 cm2 ja 1 mm2. Nendekeelte lineaarsed tihedused on siis vastavalt 0.59 kg/m ja 0.0059kg/m ning soovitatav pinge 1.1 · 105 N ja 1.1 · 103 N.Kiirus avaldub kujul

√Tρ =

√1.1·105 N0.59 kg/m = 432

ms

Kui pikk peab olema keel, mis vastab 1. oktaavi do-le (262Hz)?

262 Hz = 12l432ms ⇒ l = 0.82 m


Ülesanne

Klaveri keeled on valmistatud terasest tihedusega 5900 kg/m3.Tootjad soovitavad rakendada pinget ligikaudu 1.1 · 109 N/m2.Milline on helilaine levimise kiirus selles keeles? Kas see sõltubristlõike pindalast?

Vaatame kahte keelt ristlõike pindaladega 1 cm2 ja 1 mm2. Nendekeelte lineaarsed tihedused on siis vastavalt 0.59 kg/m ja 0.0059kg/m ning soovitatav pinge 1.1 · 105 N ja 1.1 · 103 N.Kiirus avaldub kujul

√Tρ =

√1.1·105 N0.59 kg/m = 432

ms

Kui pikk peab olema keel, mis vastab 1. oktaavi do-le (262Hz)?

262 Hz = 12l · 432ms ⇒ l = 0.82 m


Ülesanne

Miks on klaveril eksponentfunktsiooni kuju?


Ülesanne

Oletame, et kõik keeled on samast materjalist, sama ristlõikepindalaga ning sama pinge all (andmed eelmisest ülesandest).Valemi ν = 1

2l· 432 abil saab välja arvutada kõigi klaverikeelte

pikkused.

Klaveril on 88 klahvi, kõige madalama heli, subkontra oktaavila sagedus on 27,5 Hz. Seega i-nda keele pikkus li rahuldabseost

27,5 · ( 12√

2)i−1 =1

2li· 432, i = 1, . . . , 88.

Teisendades saame

li =7,85

( 12√

2)i−1, i = 1, . . . , 88.


Ülesanne

li =7,85

( 12√

2)i−1

l1 = 7.85

l40 = 0.825

l88 = 0.052


Keele algasend ja algkiirus

Olgu s0(x) = y(x, 0) keele algasend ja v0(x) algkiirus(0 6 x 6 l). D’Alembet’i lahendi abil saab tuletada keelevõnkumise võrrandile ühese lahendi

y =1

2(s0(x+ ct) + s0(x− ct)) +

1

2c

∫ x+ctx−ct

v0(u) du.

Näide. Olgu 0 < a < l ning

s0(x) =

{x/a, 0 6 x 6 a,(l − x)/(l − a), a 6 x 6 l.


Keele algasend ja algkiirus


Poognaga mängitavad keelpillid

Vaatleme poogna abil võnkuma pandud pillikeelt ningkonkreetset punkti sellel. Keele kõrvalekallet tasakaaluasendistkirjeldavad järgmised joonised.

Keele kõrvalekalle poognale lähedases punktis:

3.4. THE BOWED STRING 95

Helmholtz5 carried out experiments on bowed violins, using a vibra-tion microscope to produce Lissajous figures. He discovered that the motionof the string at every point describes a triangular pattern, but with slopeswhich depend on the point of observation. Near the bow, the displacementis as follows,

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@@

whereas nearer the bridge it looks as follows.

This means that the graph of velocity against time has the following form

where the area under the axis equals the area above, and the width of thetrough decreases towards the bridge.

The interpretation of this motion is that the bowing action alternatesbetween two distinct phases. In one phase, the bow sticks to the string andpulls it with it. In the other phase, the bow slides against the string. Thisform of motion reflects the fact that the coefficient of static friction is higherthan the coefficient of dynamic friction.

The resulting motion of the entire string has the following form. Theenvelope of the motion is described by two parabolas, a lower one and an in-verted upper one. Inside this envelope, at any point of time the string hastwo straight segments from the two ends to a point on the envelope. Thispoint circulates around the envelope as follows.

�

-

?

bow

bridge

To understand this behaviour mathematically, we must solve the fol-lowing problem. What are the solutions to the wave equation (3.2.2) satis-fying not only the boundary conditions y = 0 for x = 0 and for x = ℓ, forall values of t, but also the condition that the value of y as a function of t

5See section V.4 of [51].

Keele kõrvalekalle roobile lähedases punktis:



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@@��

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@@






�

-

?

bow

bridge


5See section V.4 of [51].


Poognaga mängitavad keelpillid

Kogu keele võnkumist kirjeldab järgmine joonis



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@@






�

-

?

bow

bridge


5See section V.4 of [51].Evely Leetma Muusikainstrumentide füüsikalisi aspekte 16 / 25

Puhkpillid

Puhkpillide korral vaadeldakse kahte muutujat - nihet jaakustilist rõhku. Mõlemad rahuldavad lainevõrrandit.

Rajatingimused sõltuvad sellest, kas tegemist on kinnise võilahtise toru otsaga. Kinnise toru otsas on õhuosakeste nihenull, lahtise toru otsas rõhk null.100 3. A MATHEMATICIAN’S GUIDE TO THE ORCHESTRA

pressure displacement

Bear in mind that the vertical axis in this diagram actually representshorizontal displacement or pressure, and not vertical, because of the longitu-dinal nature of air waves. Furthermore, the two parts of the graphs only rep-resent the two extremes of the motion. In these diagrams, the nodes of thepressure diagram correspond to the antinodes of the displacement diagramand vice versa. The second and third vibrational modes will be representedby the following diagrams.

Tubes or pipes which are closed at one end behave differently, becausethe displacement is forced to be zero at the closed end. So the first two modesare as follows. In these diagrams, the left end of the tube is closed.


It follows that for closed tubes, odd multiples of the fundamental fre-quency dominate. For example, as mentioned above, the flute is an opentube, so all multiples of the fundamental are present. The clarinet is a closedtube, so odd multiples predominate.

Conical tubes are equivalent to open tubes of the same length, as il-lustrated by the following diagrams. These diagrams are obtained from theones for the open tube, by squashing down one end.

100 3. A MATHEMATICIAN’S GUIDE TO THE ORCHESTRA








Puhkpillid

Lahtiste torude korral (nt flööt) esinevad kõik harmoonikud,kinniste torude korral (nt klarnet) domineerivad põhisagedusepaarituarvulised kordsed.









Puhkpillid

Kooniliste torude korral (nt oboe) esinevad kõik harmoonikud3.5. WIND INSTRUMENTS 101


The oboe has a conical bore so again all multiples are present. Thisexplains why the flute and oboe overblow at the octave, while the clarinetoverblows at an octave plus a perfect fifth, which represents tripling the fre-quency. The odd multiples of the fundamental frequency dominate for a clar-inet, although in practice there are small amplitudes present for the evenones from four times the fundamental upwards as well.

At this point, it should be mentioned that for an open end, p = 0 isreally only an approximation, because the volume of air just outside of thetube is not infinite. A good way to adjust to make a more accurate represen-tation of an actual tube is to work in terms of an effective length, and con-sider the tube to end a little beyond where it really does. The following di-agram shows the effective length for the fundamental vibrational mode of aflute, with all holes closed.

The end correction is the amount by which the effective length exceeds theactual length, and under normal conditions it is usually somewhere aroundthree fifths of the width of the tube.

The effect of an open hole is to decrease the effective length of the tube.Here is a diagram of the first vibrational mode with one open hole.


Puhkpillid

Flöödi augu avamine on samaväärne toru lühemakstegemisega.

3.5. WIND INSTRUMENTS 101


The oboe has a conical bore so again all multiples are present. Thisexplains why the flute and oboe overblow at the octave, while the clarinetoverblows at an octave plus a perfect fifth, which represents tripling the fre-quency. The odd multiples of the fundamental frequency dominate for a clar-inet, although in practice there are small amplitudes present for the evenones from four times the fundamental upwards as well.

At this point, it should be mentioned that for an open end, p = 0 isreally only an approximation, because the volume of air just outside of thetube is not infinite. A good way to adjust to make a more accurate represen-tation of an actual tube is to work in terms of an effective length, and con-sider the tube to end a little beyond where it really does. The following di-agram shows the effective length for the fundamental vibrational mode of aflute, with all holes closed.

The end correction is the amount by which the effective length exceeds theactual length, and under normal conditions it is usually somewhere aroundthree fifths of the width of the tube.

The effect of an open hole is to decrease the effective length of the tube.Here is a diagram of the first vibrational mode with one open hole.


Membraani võnkumine

Vaatleme ümarat trummi. Trumminaha tihedus olgu ρ (masspindalaühiku kohta) ja pinge rajal T (njuutonit meetri kohta)st kogu pind on ühtlase pinge all.

Asugu membraan xy-tasapinnal. Membraani võnkumisevõrrand on järgmine

∂2z

∂t2= c2

(∂2z

∂x2+∂2z

∂y2

),

kus c =√T/ρ ning z(x, y, t) on membraani kõrvalekalle

tasakaaluasendist ajal t punktis (x, y).



Minnes üle polaarkoordinaatidele (r, θ), saame võrrandi

∂2z

∂t2= c2

(∂2z

∂r2+

1

r

∂z

∂r+

1

r2∂2z

∂θ2

),

mille üldlahend on

z = AJn(wr/c) sin(wt+ φ) sin(nθ + ψ),

kus A, w, φ, n ja ψ on mingid konstandid, Jn Besselifunktsioonid.



Statsionaarsed punktid ja suhtelised sagedused


We have seen that to choose a vibrational mode, we must choose a nonneg-ative integer n and we must choose a zero of Jn(z). Denoting the kth zeroof Jn by jn,k, the corresponding vibrational mode has frequency (cjn,k/2πa),which is jn,k/j0,1 times the fundamental frequency. The stationary pointshave the following pictures. Underneath each picture, we have recorded thevalue of jn,k/j0,1 for the relative frequency.

&%'$

1.0000

&%'$

1.5933

&%'$

2.1355

&%'$

��

TT

TT

2.6531

&%'$�

��

@@

@

3.1555

&%'$

��2.2954

&%'$

��2.9173

&%'$

��3.5001

&%'$��

TT

TT

4.0589

&%'$��

��

@@

@

4.6010

&%'$

��j3.5985

&%'$��n4.2304

&%'$��

4.8319

&%'$��

TT

TT

5.4121

&%'$��@@@5.9765

n = 0 n = 1 n = 2 n = 3 n = 4

k = 1

k = 2

k = 3

In the late eighteenth century, Chladni6 discovered a way to see nor-mal modes of vibration. He was interested in the vibration of plates, butthe same technique can be used for drums and other instruments. He placedsand on the plate and then set it vibrating in one of its normal modes, us-ing a violin bow. The sand collects on the stationary lines and gives a pic-ture similar to the ones described above for the drum. A picture of Chladnipatterns on a kettledrum can be found on page 107.

In practice, for a drum in which the air is confined (such as a kettle-drum) the fundamental mode of the drum is heavily damped, because it in-volves compression and expansion of the air enclosed in the drum. So whatis heard as the fundamental is really the mode with n = 1, k = 1, namely thesecond entry in the top row in the above diagram. The higher modes mostlyinvolve moving the air from side to side. The inertia of the air has the effectof raising the frequency of the modes with n = 0, especially the fundamen-tal, while the modes with n > 0 are lowered in frequency in such a way asto widen the frequency gaps. For an open drum, on the other hand, all thevibrational frequencies are lowered by the inertia of the air, but the ones oflower frequency are lowered the most.

The design of the orchestral kettledrum carefully utilises the inertia ofthe air to arrange for the modes with n = 1, k = 1 and n = 2, k = 1 to have

6E. F. F. Chladni, Entdeckungen über die Theorie des Klanges, Weidmanns Erben undReich, Leipzig, 1787.



3.6. THE DRUM 107

Chladni patterns on a kettledrumfrom Risset, Les instruments de l’orchestreEvely Leetma Muusikainstrumentide füüsikalisi aspekte 24 / 25

Idiofonid120 3. A MATHEMATICIAN’S GUIDE TO THE ORCHESTRA

λ1

λ2

λ3

etc.

For actual instruments, rather than the idealized bar described above,the series of partials is somewhat different. Tubular bells are the closest tothe ideal situation described above, with second and third partials at fre-quency ratios of 2.76:1 and 5.40:1 to the fundamental.

The bars of an orchestral xylophone are made of rosewood, or some-times of more modern materials which are more durable and keep their pitchunder more extreme conditions. There is a shallow arch cut out from the un-derside, with the intention of producing frequency ratios of 3:1 and 6:1 forthe second and third partials with respect to the fundamental. These par-tials correspond to tones an octave and a perfect fifth, respectively two oc-taves and a perfect fifth above the fundamental.

a aThe marimba is also made of rosewood, and the vibe is made from alu-

minium. For these instruments, a deeper arch is cut out from the underside,with the intention of producing frequency ratios of 4:1 and (usually) 10:1with respect to the fundamental. These represent tones two octaves, respec-tively three octaves and a major third above the fundamental.

a aThe tuning of the second partial can be made quite precise, because ma-

terial removed from different parts of the bar affect different partials. Remov-ing material from the end increases the fundamental and the partials. Tak-ing material away from the sides of the arch lowers the second partial, while

3.12. THE BELL 129

A bell can be thought of as a very deformed plate; its vibrational modesare similar in nature, but starting with n = 2. But the exact shape of the bellis made so as to tune the various vibrational modes relative to each other.There are five modes with special names, which are as follows. The mode(n, k) = (2, 0) is the fundamental, and is called the hum. The prime is (2, 1),and is tuned to twice the frequency, putting it an octave higher. There aretwo different modes (3, 1), one of which has the stationary circle around thewaist, and the other nearer the rim. The one with the waist is called thetierce, and is tuned a minor third above the prime. The other mode, some-times denoted (3, 1♯), with the stationary circle nearer the rim is called thequint. It is pitched a perfect fifth above the prime. The nominal mode is(4, 1), tuned an octave above the prime, so that it is two octaves above thehum. The nominal mode is by far the one with the largest amplitude, sothat this is the perceived pitch of the bell. Mode (4, 1♯) is sometimes calledthe deciem, and is usually tuned a major third above the nominal. It canbe imagined that a great deal of skill goes into the tuning of the vibrationalmodes of a bell. It is an art which has developed over many centuries. Par-ticular attention is given to the construction of the thick ring near the rim.The information described above is summarized in the following diagram.

1:1Hum

(2, 0)

2:1Prime

(2, 1)

12:5Tierce

(3, 1)

3:1Quint

(3, 1♯)

4:1Nominal

(4, 1)

5:1Deciem

(4, 1♯)

The singing bowl. Our discussion of the bell applies equally well to otherobjects such at the Tibetan singing bowl, which is used mainly for ritual pur-poses. The singing bowl may be struck with the wooden mallet on the in-side of the rim to set it vibrating, or it may be stroked around the outsideof the rim with the mallet to sustain a vibration in the manner of stroking awineglass. After the mallet is removed, the tone can often last in excess of aminute before it is inaudible.


Muusikainstrumentide füüsikalisi aspekte · 2007. 4. 11. · Pingul keele seadus (law of...

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