# Dispersion Relation: Beads and String

1. Mar 19, 2013

### mathskier

When making the transition from the dispersion relation for a beaded string to the relation for a continuous string, I'm confused about the following issue. Take a to be the spacing between beads, m the mass of each bead, and T the tension in the string. We assume these to be constant.

For the beaded string, we have that $\omega_{n}=2 \omega_{o} \sin(\frac{k_{n} a}{2}).$ Clearly, the value of $\omega_{n}$ can never exceed $2 \omega_{o}$.

When we transition to a continuous string, by taking the limit as $a \rightarrow 0$, we end up seeing that $\omega_{n} = \frac{n \pi}{L} \sqrt{\frac{T}{\mu}}$. But if we let $n$ get very, very large, this should be able to surpass $2 \omega_{o}$, assuming we allow n to become large enough. So why is it that if we had a billion (or trillion or googol...) masses on a string there would be no way for the frequency of the highest-frequency normal mode to exceed a certain value, but as soon as we take the continuum limit, the frequencies can no get arbitrarily large?

Last edited: Mar 19, 2013
2. Mar 20, 2013

### mathskier

Whoops, meant to say "can *not* get" at the end!

3. Mar 20, 2013

### klawlor419

I think the answer is that the continuum limit is only valid when the argument in the Sin function in your dispersion relation is much less than 1, or simply close to zero. In that case you can Taylor expand and get to the relation that you have. As n, the mode number gets larger and larger the approximation gets worse. As the mode number increases the wavelength of standing wave decreases and gets closer and closer to the continuum spacing of the beads, this is probably bad terminology but its sort of physically what cause the approximation to break down.

My point is that if n gets really large in the continuum limit case, the value of the frequency it gives you is not accurate anyways so it doesn't really matter that it is more than the 2(zero mode frequency).

4. Mar 20, 2013

### mathskier

But presumably we should be able to take the number of beads (in the denominator of k) to be large enough so that the continuum limit holds up far enough for us to at least find a frequency over twice the natural frequency (omega-naught).

5. Mar 20, 2013

### klawlor419

So I see the continuum limit as being n the number of modes must be much less than than number of particles, or more accurately the number of spacings between the particles. Meaning that if you change the number of modes so they approach the number of particles or beads the continuum limit is not valid. So there is no way you can get a frequency bigger than the 2(natural freq). What are you using for the value of kn?

6. Mar 20, 2013

### mathskier

Let's call N the number of beads, n the mode number, a the spacing. kn=(n Pi)/(N+1)... so the continuum limit could be maintained by making a small enough and N large enough, and then going to a very high numbered mode, correct? For example, let's say you go to the 100th mode of the continuous string. How does that remain below 2*omega-naught?

7. Mar 20, 2013

### klawlor419

I could be just thinking about this the complete wrong way. But attached is the argument I have in mind.

#### Attached Files:

• ###### standingwavestring.pdf
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8. Mar 20, 2013

### mathskier

What you post makes sense! I guess that thinking about the limiting process might be problematic. But we can certainly think of a string in some normal mode (ie executing a standing wave) with very high mode number. The frequency is necessarily very large, as given by n*Pi/L. I'm worried about why those large mode numbers seem to approach a very high frequency for a continuous string, when anything not continuous is limited.

9. Mar 21, 2013

### klawlor419

It makes sense that its not well defined in that case. I mean if the number of modes is exactly the number of particles, each particle oscillates up and down.. no longer a continuum (a large segment of the string) oscillating collectively. If you want to explore those higher n modes, then you have to use the beaded string approach, it would be a hell of a computational problem but that it how you would do it, exactly.

10. Mar 21, 2013

### olivermsun

I think what you guys have found is the Nyquist theorem.

11. Mar 21, 2013

### mathskier

Well my point is that a continuous string has an infinite number of modes. Why are the frequencies (as in, the modes) not limited for a continuous string?

12. Mar 21, 2013

### olivermsun

Maybe go back and rewrite the dispersion relationship, but write out your $\omega_0$ explicitly. That will give a hint as to why the frequencies can get very high as you approximate a continuous string by dividing it into ever smaller segments.

Physically, there are unlimited frequencies in the continuous string because there are infinitely many sinusoidal functions which satisfy the boundary conditions at the clamped ends, and the frequency is just the wavenumber times the wave phase speed (how many wiggles travel past a point per time).

13. Mar 21, 2013

### mathskier

Aha! The omega-naught has an a in the denominator, so is that not staying constant as we take a continuum limit? In fact, is it approaching infinity itself?

14. Mar 21, 2013

### olivermsun

Yeah, you aren't just adding more and more small segments (thereby increasing the total mass + length of the string), you're dividing the string into tinier masses closer together.

(Edit: I see now you were talking about a, but there should also be an m in there).

Last edited: Mar 21, 2013
15. Mar 22, 2013

### mathskier

But both a and m are in the denominator, so that reasoning can still be asked about.

16. Mar 22, 2013

### olivermsun

What happens to $1/\sqrt{ma}$ as both $m,a \rightarrow 0$?

17. Mar 22, 2013

### mathskier

It goes to infinity?

18. Mar 23, 2013

### olivermsun

19. Mar 25, 2013

### klawlor419

Oh wow I completely missed that extra factor of a in the denominator. Fortunately expanding that argument Sin[const. a]/a about a->0 is doable.. see the attached for my correction. Still not sure if your question is answered so let me know

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