Energy losses in a vibrating string


by cesiumfrog
Tags: energy, losses, string, vibrating
cesiumfrog
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#1
Oct27-10, 09:35 PM
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We recently had a thread on energy losses in a pendulum. What about a string? After it is randomly plucked, what model explains why the many very high order modes decay so much quicker than the fundamental resonance mode does?

A similar example (if this might be any easier to model) is how an air cavity such as a seashell, excited by the surrounding noise, preferentially absorbs energy from components of that noise other than near the cavity's several lowest natural frequencies?
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Andy Resnick
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#2
Oct27-10, 10:06 PM
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I don't think it's anything other than simple resonant behavior:

http://en.wikipedia.org/wiki/File:Resonance.PNG

When you drive system off-resonance, the energy quickly dissipates. I was hoping to find a simple relationship between the 'transmissibility' (from the diagram) and the decay time, and came up with this:

http://www.physicsforums.com/showthread.php?t=320839

(Post #4)
cesiumfrog
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#3
Oct28-10, 05:57 AM
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How does it dissipate?

Andy Resnick
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#4
Oct28-10, 07:41 AM
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Energy losses in a vibrating string


For the string, friction at the attachment points. For the seashell, viscous dissipation.
cesiumfrog
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#5
Oct28-10, 04:24 PM
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Then why exactly does that form of disspitation proceed at a slower rate if the system is on resonance?
Andy Resnick
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#6
Oct28-10, 06:08 PM
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If I understand your question (which is perfectly reasonable), then the answer is that we don't yet have a microscopic theory for dissipative processes.

The damping rates are determined experimentally- here's two examples:

Cavity ring-down spectroscopy
http://en.wikipedia.org/wiki/Cavity_...n_spectroscopy

Vibration testing of objects:
http://www.google.com/url?sa=t&sourc...pOK_-mH3L0hvjg

In the simple model of a resonant system, either the damping rate is specified to generate the width of the resonance peak or the width of the peak is specified to calculate the efficiency of pumping the system off-resonance. Complicated systems have multiple resonances, and can have multiple damping rates:

http://www.google.com/url?sa=t&sourc...MZg09i1qAFW_mA

But again, fundamentally, there is no microscopic theory for dissipation, and no first-principles calculation regarding the rate of energy loss via dissipation, AFAIK- maybe for a simple two-level system, like dressed states of an atom + microcavity.
cesiumfrog
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#7
Oct28-10, 07:35 PM
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Here are some high speed videos of a plucked string physics demonstration. For example, ALE-30-150-01-2.mov: the initial part shows that the pluck causes a starkly triangular travelling wave. However, the ending shows this developing into a standing wave.

What terms would I need in a computer model to approximate the complete motion of this string?
  • Air resistance (say, proportional to a power of transverse velocity) on the sections of the string itself? Or is this negligible like was said in the thread on modelling energy losses in a pendulum? (I've yet to see a plucked string in vacuum..)
  • Bending friction in the string? (I personally expect not, and also not nonlinearities in the wave function, on the basis that I think the evolution will still approach a standing wave regardless of how small the initial amplitude is.)
  • Is damping by the support structure more important? If so, this damping reaction force should surely be a simple function of the forces exerted on it by the end of the string, right? (Essentially the support is not exactly rigid so the end of the string is permitted to perform some work.) Particularly, its functional form should not encode the specific resonant frequency (which depends on the length of the string and not just the properties in the support), so I would consider it enlightening if this interaction alone is responsible for the development toward a standing wave.

Andy, I don't know what you envision "a microscopic theory for dissipative processes" might be like (and how it could differ from current knowledge), but regardless, I don't see why any microscopic theory should be necessary to completely explain such a large scale classical demonstration as the one exemplified in that video?
Andy Resnick
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#8
Oct28-10, 10:00 PM
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Maybe I didn't understand what you are asking. Now it seems like you are only asking for a time-dependent model of a resonant system.

One simple approach is to model the string as a having a single resonant frequency (given by the length, tension, mass density, etc.) If the string is fixed at each end, the resonant mode is a pure sinusoid (or e^i(kx))

Now, when you pluck the string, the triangle shape (if that's what you want to start with) is a sum of sines and cosines- for a simple triangle function, the expansion is found here for a time series:

http://en.wikipedia.org/wiki/Triangle_wave

Your expansion is in terms of length (kx/L), though, not time (wt). Using an off-centered triangle function changes the coefficients, but the different amplitudes are fairly straightforward to calculate, and many programs will do it for you.

Now you need a dispersion relation- a way to relate the spatial wavelength to a temporal frequency. A simple one will do- for a stretched string it's here:

http://en.wikipedia.org/wiki/Dispers...es_on_a_string

That's the input. Now you multiply the Fourier series with the frequency response of the string. This is a basic method of linear systems analysis. Because there is only one resonant frequency, the frequency response of the string uses a (complex) Lorentzian function. You must put in the width of the resonant peak by hand- this is the same thing as saying you must put in, by hand, the dissipation.

To summarize so far, you start off with a string shape given as a Fourier series A_k * e^i(kx/L), convert it to A_w e^i(wt) by the dispersion relation, and multiply by the lineshape to give something like A_w e^i((2w-w_0) + ig(w-w_0))t, where w_0 is the resonance frequency and g the damping coefficient. So it's clear that off-resonant modes decay as A_w *e^(-g(w-w_0)t). I may have left out some factors of '2' and 'pi'.

All my reference books are in my office- Seigmann has a particularly clear derivation in 'Lasers"; I can check the specifics tomorrow.

Does this help?
cesiumfrog
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#9
Oct28-10, 10:57 PM
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What I'm trying to do is to not input the frequency response ad hoc as you seem to be describing, but instead to see how the frequency response emerges from the classical Newtonian dynamics.

To put it another way: consider some system (like a taut string of particular length), and we can identify (ideally) the infinitely many (but countable) resonant modes. Now, most any arbitrary physically reasonable signal can be expressed as a weighted sum of only (infinitely many) such modes. But physically we know that a random signal (like a pulse or white noise) will tend to preferentially excite the fundamental (or other low order modes). This is kind of odd if you think about it because, if it were an ideal resonator, it should be equally possible to excite any higher harmonic (or a superposition of harmonics, which could look just like the original signal). So obviously in real-life resonators, energy dissipates quicker from higher harmonics than lower ones. (This observation seems to be equivalent to observing the shape of the typical real-world greens function, or related as you say, to the real-world spectral response curves.) But my question is why do real resonators consistently have this property? What is it physically that causes their behaviour to differ from the simplest idealised resonators?
Andy Resnick
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#10
Oct29-10, 09:01 AM
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Quote Quote by cesiumfrog View Post
What I'm trying to do is to not input the frequency response ad hoc as you seem to be describing, but instead to see how the frequency response emerges from the classical Newtonian dynamics.
I think I understand your question a little better. Each resonance has a different damping coefficient. As for "why", again, we have no theory of dissipation. It's put in by hand.

There's a good discussion starting on page 40 of this:

http://books.google.com/books?id=9CR...rating&f=false

Morse also wrote a good book on the subject-

http://asa.aip.org/books/vib.html

Bottom line, the basic model is fairly straightforward to code- it's just a damped oscillator. Note, if you supply a constant power noise signal, the resonant behavior is not *amplified*- you are constantly supplying power. *All* the resonances are excited- this is why the sound of air moving across a tube is not a pure tone.
cesiumfrog
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#11
Nov4-10, 10:54 PM
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Quote Quote by Andy Resnick View Post
If I understand your question (which is perfectly reasonable), then the answer is that we don't yet have a microscopic theory for dissipative processes.

The damping rates are determined experimentally[...]

But again, fundamentally, there is no microscopic theory for dissipation, and no first-principles calculation regarding the rate of energy loss via dissipation, AFAIK- maybe for a simple two-level system, like dressed states of an atom + microcavity.
Quote Quote by Andy Resnick View Post
Each resonance has a different damping coefficient. As for "why", again, we have no theory of dissipation. It's put in by hand.
I don't understand what you mean about needing a "microscopic" theory for dissipative processes. Isn't it a straight forward matter to work out the viscous drag near the walls of a cylindrical tube, and thereby calculate "from first principles" the response spectra of an instrument such as a Helmholtz resonator (or a seashell)? Isn't the drag on a vibrating metal wire a straightforward function of the transverse velocity of that section of wire, sufficient for us to calculate the way a plucked steel-string instrument will degenerate to a fundamental resonance (and conclude an ideal string in vacuum would not)?

If you accept this, then what I think is more interesting is that I suspect the same ultimate general behaviour (always preferentially exciting lower harmonics for the longest, regardless of the spectra of the original driving noise signal) does not depend on the specific mechanism of the dissipation (e.g, whether it is due to viscosity or nonelasticity or nonrigidity, so if we took a nearly-ideal string into strong vacuum I would expect similar behaviour to be produced through an entirely different mechanism) and I'm interested in understanding the crucial features of this. I would guess that it has to do with superlinear velocity dependence.
Andy Resnick
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#12
Nov5-10, 09:01 AM
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Cesiumfrog,

I wonder if we are talking past each other. Clearly, we understand the dynamical behavior of a driven damped oscillator. We also understand the dynamics of a coupled system of driven damped oscillators:

http://www.icpress.co.uk/etextbook/p352/p352_chap1.pdf
http://iopscience.iop.org/1751-8121/..._10_105302.pdf
http://www.jneurosci.org/cgi/content/full/29/29/9351
http://books.google.com/books?id=6Pv...upling&f=false

My point centers on the assignment of a value of the damping coefficient- yes, there are models (e.g. relaxation times), but again these are phenomenological models- the damping has to be measured.

Let's take your example- let's determine the effect of viscous drag on the vibrating string:

That's straightforward to estimate. The drag on a cylinder per unit length, if the cylinder moves perpendicular to the axis and there is no turbulent flow, is given by

[tex] f_{drag}(z) = \frac{4\pi\eta v(z)}{1/2 - \gamma-ln(\frac{a\rho v(z)}{4\eta})} [/tex]

where [itex]\eta[/itex] is the kinematic viscosity of the fluid, [tex]\gamma [/itex] is Euler's number, [tex]\rho[/itex] the fluid density, a the cylinder radius, and v the fluid velocity.

For a guitar string in air, take a = 1 mm, the usual values of viscosity and density for air, and estimate the string velocity as 2 m/s (4mm amplitude at 500 Hz), a 1m string experiences a viscous drag of about 2*10^-4 N. If the approximation of low Reynolds number is valid...

The corresponding Reynolds number is about 100; so we can use the formula for drag.

Ok, so I rough calculated the drag force- can I estimate the damping time? Yes- the damping coefficient is given by b= f/v (from dimensional analysis), and so the damping coefficient divided by the mass of the string gives 1/T, where T is the relaxation time.

Note- the damping coefficient is (nearly) velocity *independent* for viscous drag.

Using the mass of a string as 20 g- I have no idea how much they weigh- gives a relaxation time of 200 seconds- this is obviously too long. Also, the relaxation time is independent of the vibrational mode of the string- the damping coefficient is independent of velocity- so it's not clear how certain modes have different relaxation times.

One approach is to introduce further complexity- take into account how the string is mounted at the ends, account for the resonant coupling between string and guitar body, energy losses due to the pickups (for an electric guitar), etc. etc. Clearly, this quickly makes the problem intractable.

Another approach is to simply measure the decay time(s) with a spectrum analyzer and then use that for a damping coefficient(s).

Note, the method used above was very general- the drag force is an analytic result, the conversion of drag to damping and relaxation time also analytic and independent of the specific form of the dissipation. The take-home message here is that a first-principles calculation of the system response for a driven dissipative system requires 'extra' information- the damping. Damping coefficients, except for a very select few systems (IIRC, the viscosity of liquid Argon has been calculated), cannot be calculated from first principles.

Edit: alternatively, I could say we don't have a way to calculate the (complex) susceptibility from first principles.

Does this make sense?


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