6
$\begingroup$

In this lecture on Forced Oscillations, Normal Modes, Resonances, Musical Instruments, the professor says that by moving a bow over a violin string, you expose it to a lot of frequencies.

Is there a more precise explanation? I mean, when I move a bow in one direction orthogonally to a string, by friction a force is exerted on the string. So, when this force equals the force induced by the tension of the string, then an equilibrium is created and a displacement of the string results. No oscillation. So, where does these "infinitely many frequencies" come from?

Later, the professor moves a tube in the air. However, here the motion is circular and so the direction of force exerted by the wind, at the top opening of the tube, will indeed pass by in an ocillating motion. So, this is a different excitation that the bow moving orthogonally in one direction.

Also later, the professor bangs a tuning fork. In my opininion, banging is not a forced oscillation at all, it is a normal harmonic oscillation, where the bang are the initial conditions.

Anyway, but what baffles me the most is the explanation of the violin string. So, could anybody explain more precisely why the bow "exposes it to infinitely many frequencies"?

$\endgroup$
4
  • $\begingroup$ An 'excited' tuning fork is not a harmonic oscillator because it has multiple modes of oscillation. $\endgroup$ – Gert Jan 31 at 15:28
  • $\begingroup$ Violin and bow: en.wikipedia.org/wiki/…. $\endgroup$ – Gert Jan 31 at 15:30
  • $\begingroup$ Tuning fork modes: physics.stackexchange.com/questions/266008/… $\endgroup$ – Gert Jan 31 at 15:33
  • $\begingroup$ @Gert Okay, thank you. The interaction between the bow and the string is more a steady "stick-slip phenomenon". Yes, the fork is not a harmonic oscillator, but banging it is also not a forced oscillation I would say... $\endgroup$ – StefanH Jan 31 at 15:39
10
$\begingroup$

In musical instruments, none of the excitations, whether by bowing, plucking, reed vibration, lip vibration, striking, etc, are sinusoidal methods. Therefore, they all, by Fourier's theorem are composed of multiple sine wave frequencies. If one or more of these frequencies corresponds to resonant frequencies of the instrument, the energy of those components will persist longer than other frequencies. If energy continues to be input to the instrument, the resonant frequencies will persist, giving rise to the continuing musical tone.

Even a brief strike, like a mallet blow, has a time behavior, and resembles either a square, sawtooth, or triangle pulse, can be viewed as a pulse of multiple frequencies. It need not be a multiple period train. A single pulse induces the forcing function. Whether it affects the object depends on the damping of the object. Consider the striking of a tympani (low damping) versus a snare drum (higher damping).

Some excitation methods, especially brass-instrument mouthpieces and reeds, begin the tone with multiple frequencies, then the reflected wave in the instrument creates a feedback loop, locking the lips or reed into resonant frequencies of the instrument. This is especially important for brass players, who can practice with a mouthpiece only playing a tune close to the actual pitches of the music. That way, they are adjusted to optimize the frequency feedback.

The violin (and others) with bowing action is excited by a "grab-release" action of the resin-coated horse hair on the string. The action very strongly resembles a continual sawtooth wave plucking action. The string's length, tension, material, and thickness determine the resonant frequencies of the string, while the sawtooth wave, by Fourier's theorem, inputs more than enough frequencies to the string. The ones which match the resonant frequencies persist, while the ones which don't match dampen very quickly.

$\endgroup$
2
  • $\begingroup$ Thank you. To make this back and forth "sawtooth" motion, it is essential that the frictional coefficient for a moving object (in this case the bow) is smaller than for a not moving object (so that it can slip back), right? $\endgroup$ – StefanH Jan 31 at 16:40
  • $\begingroup$ @StefanH That's true in general: kinetic coeffiients are <= static coefficients, on the other hand, the force of the hand pushing and pulling the bow can be increased or decreased independent of the friction. The net force on the bow determines its acceleration. $\endgroup$ – Bill N Jan 31 at 18:14
5
$\begingroup$

So for the more precise explanation, you have a harmonic oscillator with some sort of drag term and some sort of force, $$ \frac{\mathrm d^2x}{\mathrm dt^2} = -2\lambda \frac{\mathrm dx}{\mathrm dt} - \omega_0^2 x(t) +F(t). $$ Exercise 1: prove that if $F(t)=0$ then this oscillates at frequency $\sqrt{\omega_0^2-\lambda^2}$ as it decays like $e^{-\lambda t}.$

Exercise 2: prove that if you drive with a periodic force $F(t) = A e^{i\Omega t}$ (or use sines and cosines if you are not yet comfortable with complex numbers as 2D rotations) then the response of the system is also periodic in the same $\Omega t.$ What do you get for the amplitude? When is that number highest?

Exercise 3: what does the differential equation look like in Fourier space? Exercise (2) drove the system with $F(\omega) = A\delta(\omega-\Omega)$ where $\delta$ is the Dirac delta-function, do you see the basic results of exercise 2 in Fourier space? If we view this result as multiplying the frequency-space force by a complex function, what does the graph of its absolute value look like? Especially, when the quality factor $Q=\omega_0/(2\lambda)$ is large, what do you see this graph looking like? Also, from your experience of guitars and how long a string “rings out” versus what frequency it plays, what sort of $Q$-factor is reasonable for musical instruments?

Exercise 4: what is the Fourier transform of a normal distribution with standard deviation $\sigma$, i.e. $f(x) = e^{-x^2/(2\sigma^2)}/\sqrt{2\pi \sigma^2}$? What is its width?

Once you have these basic exercises completed, maybe you can complete the explanation yourself. How I would put it is, a short sharp force is like a Gaussian with a very small standard deviation. This makes it very wide in frequency-space, which we can interpret as its having lots of different frequencies. But we saw in exercise 3 that there is this susceptibility function with a peak around $\omega_0$ and so those are the frequency parts of the driving force that the system is most going to pick up, particularly if it is a high-$Q$ resonator. Its response is in this little peak around some frequency and it muffles all of the other components to zero. (Indeed if you look up $Q$-factor in Wikipedia you will see that a common definition is the width of the resonance peak in frequency space, divided by the frequency of the peak. So this is almost the definition of a high-$Q$ resonator.)

Caveats

While I think the above answers your question, it can be somewhat dangerous to assume that we automatically know everything from knowing the littlest thing. Note that this system is very simple, not having multiple modes at various frequencies, and in many other respects it is not “real.”

Probably the easiest generalizations are finite element models, so you set up a bunch of masses along some $y$-coordinate with positions $x_i(t)$ and then their restoring forces return like $$\ddot x_i = -2\lambda \dot x_i - \omega_0^2 (2 x_i -x_{i-1} - x_{i+1})$$ so the analysis gets tied up a little with this tridiagonal matrix $$\begin{bmatrix} 2&-1&0&\cdots\\ -1&2&-1&\cdots\\ 0&-1&2&\cdots\\ \vdots&\vdots&\vdots&\ddots\end{bmatrix}$$ and the process of diagonalizing this matrix is more formally called finding the “normal modes” of the system. This “normal modes” analysis is not just the basis of mechanical engineering software, but I have seen it a lot in quantum chemistry calculations, and similar approaches can be helpful at other times, so learning differential equations that have some matrices thrown in can be very rewarding.

But you might also like the full 2d “string wave equation” which just replaces the above finite difference term with $$\frac{\partial^2y}{\partial t^2} = -2\lambda \frac{\partial y}{\partial t} + \omega_0^2\frac{\partial^2y}{\partial x^2}.$$ This launches into a discussion of wave equations more generally. In particular, you get these resonance peaks at $\omega_0,2\omega_0,3\omega_0,\dots$ but it all has to do with boundary conditions on the sides of the string, as you form standing waves on it.

If you own a guitar you may have noticed that the string wave equation assumes that the tension in the strings is held constant but actually the amplitude of real oscillations is enough to stretch the string, so the string rings out like “byoooowwww”, higher in pitch at the beginning and lowering pitch at the end. This is a fundamental nonlinearity that needs to be added to the string wave equation above. More generally a lot of linear instruments are better understood as linear-but-coupled-to-a-nonlinear-system, for which a really good article is provided by Fletcher (PDF warning).

$\endgroup$
2
  • $\begingroup$ This isn't all that relevant, because your entire analysis is fundamentally linear. The kind of phase-locking that underlies e.g. bowed strings meanwhile is a manifestly nonlinear phenomenon. $\endgroup$ – leftaroundabout Feb 1 at 16:19
  • $\begingroup$ @leftroundabout I think it suffices to give the basic point that Lewin is making about picking out one frequency out of many driving the motion, but you are right that I should cite Fletcher’s paper too. One sec. $\endgroup$ – CR Drost Feb 1 at 16:43
4
$\begingroup$

The bow pulls the string till the tension exceeds the friction force and the string skips to its initial position, to be pulled again by the bow. This happens multiple times during one complete movement of the bow. The frequency of the resulting oscillations is determined by the eigenmodes of the string (normal modes in the OP language), which are typically the integer fractions of its length. It is by controlling the length of the string that the musician changes the frequency of the sound.

This is not what I would call forced oscillations - rather we are dealing here with generating oscillations (which in itself is an important thing, but often more complex mathematically than just a forced oscillator). Forced oscillations, on the other hand, are oscillations due to periodic force, which consequently occur at the frequency of this driving force rather than with eigenfrequencies of the oscillator. Same can be said about a tube/trumpet or a tuning fork/guitar. The oscillations are created differently in all these cases, but in none of them they are created by a periodic force.

$\endgroup$
6
  • $\begingroup$ In mouthpiece instruments, the maintaining of the tone is indeed a periodic force. The initial wave train travels through the tube, reflects, and through feedback interactions with the vibrating lips sets up a resonance with the lips as the air is pushed through them. Hence, the periodic forcing function. Check out Thomas Rossings multiple books on musical instruments. $\endgroup$ – Bill N Jan 31 at 15:49
  • $\begingroup$ @Bill It is resonating at a frequency determined by the length of the pipe. Musician's lips may form a part of the resonator - which is probably what Rossling means, but no human is capable of moving their lips at frequency 440Hz - which is what one needs to drive/force the oscillations. $\endgroup$ – Roger Vadim Jan 31 at 15:54
  • $\begingroup$ Thank you. Do you know any source where I can read more about these generating oscillations? $\endgroup$ – StefanH Jan 31 at 16:00
  • $\begingroup$ Regarding the musical instruments Thomas Rossling's books, suggested by @BillN are probably a good start. Books on nonlinear theory usually deal with the subject in general but rather abstract manner. Then the books in particular fields: radio generators, laser physics, etc. $\endgroup$ – Roger Vadim Jan 31 at 16:02
  • $\begingroup$ @Vadim I encourage you to do the experiment. Get a trumpet mouthpiece, play it (or get a brass player to assist you), and do the FFT. I bet you'll find 440 and more in there. That's because the lip vibrations themselves are not sinusoidal, so the wave form of the lips will have higher frequencies tthan the lip fundamental. Also, without the continual forcing of the lips by the player, the tone will die. It is a periodic force. $\endgroup$ – Bill N Jan 31 at 16:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.