What happens when a string fixed at both ends is released from a non-mode shape?

[quasar987]

Suppose a string is fixed at both ends. What will happen if I give the string a shape that is not the shape of a normal mode, and let it go? Exemple: What happens after I give it the shape of a semicircle and let it go?

 

[inha]

hmm. I think it'll take the form of a normal mode anyway. If you do the same with a guitar for example it makes a brief off-note sound and then starts to oscillate in a normal mode giving out a proper note.

You could maybe treat this as a perturbation to a normal oscillation that goes to zero and leaves you with a normal mode as time passes.

 

[Meir Achuz]

The shape of a circle is composed of many different normal modes.
Treating it requires the use of a "Fourier transform" to give the circle shape f(x,0) in terms of sine functions as f(x,0)=Sum A_n sin(pi*n*x/L).
The constants A_n show how much of each normal mode is in the motion.
There is more math involved, but eventually you can get f(x,t) as a sum over the motion of a lot of normal modes. The moving shape of the string is complicated,
but will return to a semicircle at each end of the motion if there is no energy loss.
With energy loss (caused mostly by air resistance), the string will eventually vibrate in only the lowest mode. This usually happens so quickly that that is usually all you notice in a vibrating string. But the change from the original shape to the lowest normal mode is why a guitar goes "twang" as the high frequency modes quickly damp out.

 

[quasar987]

Thanks for the answer guys.

What is the argument that permits to conclude that any motion of the string can be expressed as a superposition of the normal modes?

 

[inha]

Do you know Fourier series? Any periodic function can be written as a Fourier series.

 

[quasar987]

Are you sure this answers it?

Cuz Fourier allows us to write the shape of the wave at a given time, or at a given point as a sum of sine and cosine, but it doesn't allow us to write the whole wave y(x,t) as

$$y(x,t)=y_1(x,t)+y_3(x,t)+y_4(x,t)+y_7(x,t)$$

for exemple (where the y_i are the normal modes). Or does it?

 

[inha]

It doesn't. Sorry. That series is basically a boundary condition y(x,0) to the wave equation you must then solve with Fourier transforms to get the general form y(x,t).

 

[quasar987]

Ok, I'm not very familiar with Fourier transforms, but as I understand you, given a wave y(x,t) with fixed ends, we can apply Fourier tranforms to find that it can be decomposed as a sum of normal modes?

 

[quasar987]

With energy loss (caused mostly by air resistance), the string will eventually vibrate in only the lowest mode.

Why do the other modes die out completely (and so rapidly!), instead of their amplitude just diminishing uniformly?

 

[inha]

Ok, I'm not very familiar with Fourier transforms, but as I understand you, given a wave y(x,t) with fixed ends, we can apply Fourier tranforms to find that it can be decomposed as a sum of normal modes?

Let's say you know the wave at t=0 y(x,0). The wave equation is a pde and you would use y(x,0) as a boundary condition to solve the wave equation with the Fourier transform method to get y(x,t). The full solution then be a series in x and t with the normal modes still present.

If you have access to Arfken,Weber: Mathematical methods for physicsts or some other equivalent book you can see how integral transforms are useful in solving PDEs. I tried to google you a link but didn't find anything that'd show in a clear manner how it's done. Sorry.

 

[Meir Achuz]

The higher modes die out faster because they are moving faster, and so have faster energy loss to the air. It's just like the energy loss due to windage increase as a car goes faster. The string is lilght, so the loss of all the energy in higher modes is rapid.
You may just have to wait til you learn more math to really understand this.

 

[quasar987]

You may just have to wait til you learn more math to really understand this.

This is the unpleasant feeling that haunts me too.

I hate the fact that all my physics classes use math that we haven't leanred yet! Was it like that for you too?

 

[HackaB]

The higher modes die out faster because they are moving faster, and so have faster energy loss to the air. It's just like the energy loss due to windage increase as a car goes faster. The string is lilght, so the loss of all the energy in higher modes is rapid.
You may just have to wait til you learn more math to really understand this.

Could you give a hint as to what math leads to this conclusion?

 

[Meir Achuz]

You don't need math for this. The faster you run, the more you find air resistance holding you back. Or just stick your hand (carefully) out a car window. The faster you go, the more force you feel on your hand. F*v s the power or the rate of energy change.

 

[HackaB]

Well, twice you've brought up the fact that "there's more math involved...", so I thought maybe there actually was. For starters, can we not model the guitar string as satisfying the wave equation, or is that too crude? If you model the air resistance with a linear damping term, all the modes still travel at the same speed (although with different frequencies, of course). And I think that means they would die out uniformly. They do the calculation at mathworld:

http://mathworld.wolfram.com/WaveEquation1-Dimensional.html near the bottom of the page.

The decay factor in the exponential is a constant. So how do you model it to account for the different mode speeds, and thus non-uniform decaying of modes, of the guitar string?

 

[Meir Achuz]

The approximation of linear damping in turbulent air is only made for textbook derivations. The twang of a guitar shows how bad an approximation it is.
Also, the only shape you can actually see (eyeball) is the one bump fundamental mode because the damping of higher modes is so fast.
Choosing (really guessing) a particular non-linear dependence on v would permit a computer solution. Higher modes oscillate up and down at a higher velocity because
v depends on n\omega_0.

 

[HackaB]

Okay. I'll buy that. So what you actually meant by

In Math Physics books, the dissipation is often described by a term
\gamma(dy/dt) in the wave equation. That is a gross simplification, but leads to simple equations for the "twang".

is that it doesn't account for twang at all, right?

 

[Meir Achuz]

Yes it doesn't. I forgot what a bad assumtion that is.