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Consider a super basic 1D vibrating string, with standing waves on it. The string has length $L$, and the wave propagates at a velocity $v$. The fundamental frequency $f_1$ is given by

$$f_1 = \frac{v}{2L}$$

And $v$ is in turn given by

$$v = \sqrt{\frac{T}{\mu}}$$

where $T$ is the tension in the string and $\mu$ is the linear mass density. There will also be standing waves at integer multiples of $f_1$.

Now, consider a vibrating loop, made of a metal (so that it tries to retain its shape), like in this video. How do I calculate the fundamental frequency of a system like this? In particular, I understand what tension means in the context of the linear system, but I don't know how it works in the loop case.

I can't see where the tension in the wire is. It's not like a guitar string being stretched taut; nobody is pulling the ring tight, it maintains its shape by itself. At first, I thought the tension would be from the string originally being straight, and then being curled into a ring, but that can't be right because the tension in the outer edge would be counter-balanced by the compression on the inner edge. And besides, you could imagine a ring being made by pouring molten metal into a very thin ring-shaped mould -- in that case there wouldn't be any stretching or compression at all -- the ring shape would be the atomic lattice's natural state. Such a ring would presumably oscillate the same way as the ring in the video.

Then, I thought that the tension comes from the wire stretching when it oscillates -- a sine wave is longer than a straight line. In this case, the tension would be related to the r.m.s. of the arc length of the wire through one oscillation(?). Apart from this requiring a horrible elliptical integral to solve, it doesn't work either, because if you increase the amplitude of the deflection you increase the tension in the wire, and that would change the fundamental frequency. But that isn't what actually happens -- you can crank the oscillator up as much as you want and it doesn't alter the position of the nodes, it just makes them easier to see (I've done this).

So perhaps that we're outside the domain the equation is valid for? My question is: is there a different, tension-like quantity I can use in the equation? Presumably it would involve some kind of bulk property of the material like the Young's Modulus, but I couldn't figure out how to make the units work out. Or a different equation entirely?

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The role of the tension in the linear string is to create a force that will pull back on any deviation from a straight line. This is what causes waves to propagate along the string. A point is moved away from equilibrium, and the tension acts to get it back to its equilibrium position. Things don't stop as the string now has kinetic energy, and keeps on moving against tension.

In a bulk material, there is already a force that will try to reduce any deviation from equilibrium. You don't need the tension, and the equation for a string is not valid. As you guessed, that force related to the elastic properties of the material. One thing that should not be confused is that in solids, you can have longitudinal waves and transverse waves, which will propagate at different speeds. There are different properties of the materials (Young's modulus, shear modulus, bulk modulus and Poisson's ratio) that must be used for each specific case. The propagation speed will also depend on the density of the material.

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Note that for thin beams (like the band loop mentioned) any tension will decrease the wavelength, and any compression will increase the wavelength. When the compression is big enough the part buckles. The source of buckling in beams is that the fundamental frequency goes to zero (wavelength becomes infinite).

To demonstrate this the fundamental wave equation in a thin beam is

$$ E I \frac{\partial^4 u}{\partial x^4} - T \frac{\partial^2 u}{\partial x^2} + \rho A \frac{\partial^2 u}{\partial t^2} =0$$

where: $u(x,t)$ is the deflection from resting shape, $T$ is the tension, $\rho$ is the mass density, $E$ is Young's modulus, $A$ is the cross-sectional area, and $I$ area moment of the section.

This is solved by the wave frequency for a beam of length $\ell$ with one end fixed by:

$$ \omega = \sqrt{ \frac{E I}{A \rho \ell^4} \Phi_i^4 - \frac{T}{A \rho \ell^2} \Phi_i^2 } $$

where $\Phi_i \approx \frac{\pi}{2} + (i-1) \pi$ is a harmonic parameter valid for $i \gg 1$. You also have $\Phi_1 = 1.87510406871196$ and $\Phi_2 = 4.69409113297417 $.

You can see that the tension $T$ causes the natural frequency to increase in a non-linear fashion. The converse statement is that when the compressive force $$T = - \frac{E I}{\ell^2} \Phi_1^2 $$ is applied the beam buckles because $\omega \rightarrow 0$.

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  • $\begingroup$ Seems there is a sign error in your differential equation--tension term should be negative, no? $\endgroup$ – Ben51 Jan 9 '18 at 16:15
  • $\begingroup$ It depends on the convention. I think you are used to seeing a negative when dealing with beams because forces are compressive (as in columns) for the interesting problems of buckling and resonance. $\endgroup$ – ja72 Jan 9 '18 at 17:18
  • $\begingroup$ I was actually thinking of strings--if you set the stiffness term to zero, you should get back the equation of motion for a string, which has the opposite sign (as it must to permit solutions of the form $u(x,t)=Ae^{i(kx-\omega t)}$. $\endgroup$ – Ben51 Jan 9 '18 at 17:30
  • $\begingroup$ You are right. Let me think about this. With the solution of $u = U \exp \left( \omega \left( \frac{x}{c}-t \right) \right)$ I think I have the signs flipped because the solution is $$ T = \frac{E I \omega^2}{c^2}- A \rho c^2$$ and for zero stiffness I get a negative tension. $\endgroup$ – ja72 Jan 9 '18 at 23:37
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If the tension on the outer rim and the compression on the inner rim would balance each other, the loop would be completely soft and you could bend it effortlessly into any shape. This is not the case, instead there is an effective tension force parallel to the surface, which pulls the ring into its resting shape (together with gravity, of course, otherwise it should form a perfect circle). You could open the ring and substitute that internal tension with an external tension force. The modes of that open band that have periodic boundary conditions should be similar to the modes of the ring. This is only a good (?) approximation for small amplitudes, of course.

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  • $\begingroup$ But wouldn't that only work if the loop was constructed by taking a straight wire and bending it? Suppose you took a sheet of metal and cut the loop out instead, so that it wasn't under any tension. Or poured liquid metal into a ring-shaped mould and let it set. Wouldn't it still retain its shape? $\endgroup$ – dain Jan 13 '16 at 1:09
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    $\begingroup$ @dain: If you take a flat metal band and you bend it, it will develop tension because of the expansion on the out rim and the compression on the inner. You can check this by letting it go, in which case it will return into the flat position (unless the material has been changed irreversibly, of course). You can also hook up a force gage to the two ends of it and measure the tension force. The same material poured into the shape would not have this tension force. $\endgroup$ – CuriousOne Jan 13 '16 at 1:27
  • $\begingroup$ Right, and presumably with such an object, you could still hook it up to an oscillator and get basically the same behavior, so how is the fundamental frequency calculated in such a scenario? $\endgroup$ – dain Jan 13 '16 at 1:34
  • $\begingroup$ @dain: As I said in my answer, as a first oder mental model you can unroll it, put the same tension on it from the outside and only consider the modes that obey the periodic conditions. If you want to calculate a simplified shape, consider a circular loop and solve the wave equation with periodic boundary conditions in polar coordinates. If you want to solve it in all generality, use the equations of continuum mechanics to formulate the full equations of motion, but that will force you to a number of approximations along the way to avoid the high frequency longitudinal modes. $\endgroup$ – CuriousOne Jan 13 '16 at 1:44
  • $\begingroup$ Stiffness causes dispersion, so modes look similar but they are not exactly at multiples of the fundamental mode. $\endgroup$ – Pieter Jun 4 '17 at 20:08

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