# Standing waves in surface vibrations in rods?

1. Jan 7, 2009

### yerpo

I appologize in advance if this problem will be awkwardly presented, but I'm a biologist, with no talent for physics or mathematics whatsoever. Here goes....

At my institution, we are researching communication of certain bugs, which use low-frequency, substrate-borne vibrations as signals. Those signals are so well tuned to the physical properties of plants the bugs live on, that they transmit with very little attenuation, and so they rebound (probably) several times from stem boundaries before dying down. As far as we know, those are surface vibrations, transmitted in the form of bending waves. I measured the amplitude of artificially-induced signals along a long straight stem of the common sedge and got a distinct pattern of amplitude minima & maxima - a sinusoid shape, repeated regularly along the whole length. We speculate (in absence of a better explanation) that the pattern represents standing waves which originate from outgoing and rebound waves adding up, but noone has been able to prove that. So I thought I'd calculate the expected amplitude from known parameters at each point and compare it to the actual data. The attempt, of course, crashed on the way through my inability to do math/physics, and I wonder if someone may point me in the right direction, at least. I'm afraid nobody has done such a thing before, though, and if this is really the case, I'd really appreciate if someone can figure it out. We could probably even arrange a co-authorship of the article for the helpful contributor (but I can't promise that 100% right now).

Anyway, to the data:

We stimulated the stem artificially with a minishaker, using pure-tone 100 Hz vibration with the velocity of 10-3 m/s (I express it with velocity because it's more relevant for the system studied, but it can easily be converted to amplitude).

The stem is circular in cross-section, around 5 mm in diameter, and 1,65 m high. It was vibrated 1 cm below the top which was cut cleanly off. I also observed equivalent pattern in shorter stems.

We don't know exactly the velocity of propagation, the best approximation I have is 95 cm/s. I don't know about the elasticity of the stem either, but I can find out if necessary. Attenuation is around 0.3 dB/cm.

There are a couple additional assumptions we made, which may of course be wrong, but it's the best we can do:
- the mass loading of the stem due to it being fixed to the minishaker doesn't significantly affect the system more than 10 cm away
- the stem's cross section is uniform enough to simplify as completely uniform and/or irrelevant to the propagation of surface vibration

The first thing I tried was the basic formula for standing waves (the one here), but I got much shorter wavelength than observed experimentally. Unfortunately, I don't know whether the formula isn't good for this purpose or if the error was mine. That's what I can think of right now. Let me know if there's anything else of importance that I left out.

Thanks.

2. Jan 7, 2009

### marcusl

This is very interesting. You gotta hand it to bugs!

This part of your story doesn't make sense: You say you think these are "surface waves...in the form of bending waves." These are different things. Bending modes are common--it's what you see if shake a sapling, for instance. Solid rods can also propagate torsional (twisting) and compressional waves. I don't know much about surface waves, I imagine they depend critically on a non-uniform cross-sectional profile (a plastic coating over a glass rod can probably carry "surface waves", or a very thin stiff coating around something soft?).

My guess is an insect would launch a longitudinal (bending) wave. Have you observed insects doing this?

Calculations for waves in solid rods are not too hard, you solve so-called "transcendental equations" for the resonance modes. The hard part will be measuring the appropriate material properties for your plant stem--Young's modulus is the main one. The other even harder part is that plants aren't simple solid bars, but are inhomogenous. The outer covering is hard and stiff, the center is softer and squishier (is that called the xylem? it's been a long time since high school biology...) and this will alter the propagation properties. Don't know whether a simplified model will suffice as you suggest.

The cadillac way to analyze an inhomogenous structure is with a big finite element simulation code, staffed by an expert who knows how to run it and interpret the results.

Hope this helps some to get you started.

Last edited: Jan 7, 2009
3. Jan 8, 2009

### yerpo

I knew I'd mess something up :P Yes, it's been proven that those are bending waves (the classical and widely cited text on this issue is Michelsen et al. 1982, although it's the assumption about standing waves made just there that we've been testing). But the way I understand it, it's the surface that bends, and the distortion is smaller the deeper you go (assuming homogeneous structure).

Like I said, the elasticity is not a problem. I looked it up now, Young's modulus is about 3*108 N/m2 for a comparable plant. It's the same order of magnitude in completely different plants.

As for the simplification, this is a herbaceous plant, without distinct stiff bark, only a green outer layer. Even so, I wish I had a more accurate model, but I'm afraid this will have to do.

So we're back at my largest problem. Many thanks for the pointer, but I wouldn't know how to even begin solving this.

Last edited: Jan 8, 2009
4. Jan 8, 2009

### marcusl

I'll take a look at the reference you provided this weekend.

5. Jan 8, 2009

### skeleton

Here are some calculations to get you started.

Part 1 of 2.

Whenever I do such calculations, I first do a simplified estimate then hope it is close to observations. If so, then I proceed with a detailed analysis; otherwise I know my model is a poor representation so I step back and revise it.

Here I have simplistically modeled the plant stem as a cantilever beam with a fixed base and a free opposite end. I would not expect the beam's (stem) response to be sensitive to any anisotropy in the sectional properties since the structural stiffness is mostly derived from the outer ring of a cylinder (note: you can compare Ig for a solid cylinder versus a hollow cylinder and notice the lack of sensitivity).

You need to measure the weight density of the beam as it is relevent. (Just weigh a very long length to factor out any local irregularities.) In the meantime I have guestimated it at 45 p/ft^3 which is probably characteristic of moist woody material (note dry wood is 35 pcf whereas pure water is 65 pcf).

I don't know the boundary conditions of your experiment. You need to identify this as it influences the "response" of the beam. I have made some assumptions in my calculations which need to be scrutinized.

Is the stem uniform in diameter over the full length? You indicated this to be so in your write-up. However, if it is instead conical (say 6 mm and tapering to 4.5 mm, then this would have a significant effect on the beam's response). You need to be accurate here.

In designing buildings structurally for seismic loading (isn't this a bit similar to your experiment?) we develop a finite element model. The model accurately determines the stiffness and response characteristics of our building tower as driven by a synthesized seismic spectrum (your forcing function). The model will determine the "participation" that each mode of vibration has upon the building. In your case, I haven't yet done this; instead, I have simplistically estimated the first 5 modes of vibration. Matching the "wave quantity" with what you observed (how many standing waves) would reveal the actual mode of vibration that has the highest participation (dominance).

Anyway, this should start you off.

#### Attached Files:

• ###### 1) Beam in vibration.jpg
File size:
25.9 KB
Views:
168
Last edited: Jan 9, 2009
6. Jan 8, 2009

### skeleton

Here are some calculations to get you started.

Part 2 of 2.

#### Attached Files:

File size:
36 KB
Views:
156
• ###### Beam corrected - 2.jpg
File size:
38.9 KB
Views:
171
Last edited: Jan 9, 2009
7. Jan 9, 2009

### yerpo

This is beginning to make sense now, thanks a lot.

I now measured my system again to remove some of the assumptions and approximations:

• d is not 5 but 4.1 mm; the beam doesn't taper towards the top significantly - I got 0.12 mm difference, but this could as well be because of me squishing it a bit.
• out of that comes a new value for Ig = 13.86 mm4
• one meter of stem weighs 9.89 g, which (using diameter 4.1 mm) comes to 734.97 N/m3, if I'm not mistaken
• beam weight is now ω = 0.09 N/m

Using your constants I got the following wave lengths:
• K = 3.52 Hz --> λ = 1583.3 m
• K = 22 Hz --> λ = 271.43 m
• K = 61.7 Hz --> λ = 96.94 m
• K = 121 Hz --> λ = 49.22 m
• K = 200 Hz --> λ = 29.69 m

The actual wavelength I observed is 20 cm, so this is way off. I suspect the wave speed is incorrect - 95 m/s (not cm as I wrote before, sorry) is one of the two numbers I got elsewhere and describes the wave speed of 200 Hz vibration in a woody branch. Would detecting the vibration at the source and some distance away, and then comparing the signal onset timing be good enough to really determine this? We use laser vibrometers for detection. Still, my result is two size categories larger, so if you'd be so kind, please re-check my calculations. Second number I got from another system is Young's modulus. How would I measure that?

One additional problem remains, assuming we get into the range of calculated wavelengths. If I just pick the right mode of vibration to suit my data, I don't prove that those are standing waves in the first place. Is there any way to tell whether I'm looking at standing waves or not? Note that I'm not really getting zero amplitude at internodes, just local minima.

8. Jan 9, 2009

### skeleton

It would help me and others if you provide a sketchs:
1) Sketch that depicts the reality of the insect and the stem, while labelling characteristic features such as boundary conditions.
2) Sketch of your test apparatus.
3) Sketch of the superimposed waves as observed.

Your modulus of elasticity, E, may be wrong. I'm a structural engineer so I only have data on lumber. The softest wood (Northern Pine) listed is E.05 = 4000 MPa. So, let's say for now that your stem is E=3000 MPa = 30*10^8 N/m^2.

E can be experimentally determined from doing a simple load-deflection test on a simple beam configuration. You apply a load at mid span of a simple beam, and measure the corresponding deflection at that point. I can later describe to you how to do this exactly and the calculations that would reveal E. But first tell me if you have laboratory equipment that can:

- Apply a light known load, say 10 gm, or increments thereof.
- Measure the resulting deflection at mid span, accurate to say 0.1 mm.

There are three types of waves:

1) Longitudinal waves. These compression wave would require very high energy in the forcing function. This can be disqualified as being infeasible for an insect.

2) Torsional waves. This requires less energy, but this would be difficult for an insect to induce.

3) Transverse waves. This is what you notice with the strings on a guitar. This takes the least energy, and would be easiest for an insect to induce if it could cause it's body to flutter at the same frequency as one of the fundamental frequencies of the plant stem. (I liken this to the head flutters of an anole lizard, or the fluttering of a cameleon as it ambulates along a branch.) It can imagine that an insect would be able to adjust its body's fluttering rate (frequency) to be sympathetic to what is developing in the plant stem (fundamental frequency); any feedback system (nervous system of the bug) could figure this out and make adjustments. Why so? Cuz once the bug adjusts its own forcing function (leg muscle twitches) to match the response frequency of the stem, then it will notice it takes far less energy to maintain the forcing function - its energy demand drops.

Note, the wave speed in a longitudinal wave is different and higher than the wave speed in a transverse wave. Perhaps you are using the longitudinal wave speed by mistake. So, you can do some further research on this datum. Additionally, as I would recommend, is to not introduce the wave speed as a parameter. Instead you can measure it. Just measure the wave length and the wave frequency; together they will reveal the wave speed. Further still, you shouldn't even care about the wave speed. You just want to confirm that you can predict the wave length from the experimental data. (Actually the wave speed can later be used to calculate the energy of the wave.)

The mode of vibration that is actually occurring corresponds to a specific energy. Putting all this data together can reveal how much work (energy/time) is being consumed by the stem to maintain its vibration. Work equals energy. So the work is being provided by the energy production of the insect as it flutters.

Another thing that would be interesting to show in your research paper is that the insect CAN produce sufficient energy to force the plant stem into vibration. Some other parameters and calculations can determine how much energy your instrument's forcing function is imparting on the plant stem. Now you need to show that the insect can potentially provide this much energy. Actually, the insect's total work is the power integrated over time, where time is the duration lapsing while the stem is increasing its amplitude.

The insect's work merely equals to the amount of energy lost from inelastic component of the vibration. How to determine this? Easy, apply a single pulse to the stem to start it vibrating, then record how long it takes for the stem to stop vibrating. Actually, you would measure the amplitudes for a set of vibrations while they are noticeable. The reduction in amplitude reveals the attenuation; this is what the insect must continuously impart onto the stem to maintain the vibration.

Last edited: Jan 9, 2009
9. Jan 9, 2009

### skeleton

Please review my original attachments; I have made corrections and incorporated your revisions. Additionally, I changed the wave velocity such that the first mode of vibration matches your observed wave length of 40 cm.

Be careful here. A wave length has one trough and one valley (up and down) cycle. This is the wavelength for which a visual standing crest or trough of 20 cm would have a wavelength of 40 cm (mathsheet reads 40.4 cm). You should confirm my understanding of your 20 cm.

If this model is correct, then you would have observed a period of vibration of 2.4 sec - very easy to notice. What frequency did you get?

Last edited: Jan 9, 2009
10. Jan 9, 2009

### skeleton

I ran another trial in pursuit of matching your results and calibrating it with expected behaviour of a cantilever.

In principle, the 1st mode of vibration of a cantilever beam has a wavelength that is 4 times the length of the cantilever. So I changed v until N=0.25. Then I looked at lambda vector to find the element that has a wavelength closest to your observed half-wavelength of 20 cm (full wavelength = 40cm). The 3rd mode of vibration reads lambda.3=36 cm. The corresponding period is 0.14 sec; that is 6 cycles a second or 1 half-cycle in 1/3 second - oh! you should be able to notice this with your eye.

All the modes are characteristic of the beam itself and its boundary conditions; the forcing function does not influence the mode vector. So, if you removed the device that previously applied the forcing function, then the beam would have boundary conditions true of a cantilever. Now you could merely tap the free end with your finger hand notice the 1st mode of vibration. The period of vibration that you measure here can be used to calibrate the mathsheet with the true wave speed.

If this revised model is correct, then you can ask why does the insect choose to over-exert itself to force the stem into a higher mode of vibration. Note, higher modes always require more energy. Well, you could rationalize that the lower modes, such as mode 1 with a period of 2.38 sec is so slow that the wind could easily achieve this and thus provide unintentional "communication signals". So, the insect deliberately is willing to exert extra energy so as to push the stem into a higher mode that would not occur during common wind conditions. This begs for another experiment: see if the insect is unaffected by lower frequency vibrations. You could easily test for what frequencies of vibration the insect is responsive to. This would then correspond to certain range of plant stems (diameter and length combinations) that they would be able to communicate on.

You have a fascinating study ...

#### Attached Files:

• ###### Beam - 4.jpg
File size:
36.7 KB
Views:
142
11. Jan 12, 2009

### yerpo

Right, let's get to the bottom of this. First, I owe some details to clarify what's happening. Some of those I deliberately withheld to not influence the calculation, but that might not have been such a good idea, because I'm beginning to get a feeling that we're not talking about the same thing.

Me and the insects induce 100 Hz vibrations in the form of pulses about a second long. Below is a graph of amplitude change in those signals with distance, expressed as maximum velocity of particle movement (= 2πf × max. displacement) at each point.

You will see that wavelength of the pattern is indeed 20 cm, but the fundamental frequency of underlying pulses (not shown here) does not change significantly and is always approx. 100 Hz. What does change is the appearance of harmonics at 200 and 300 Hz. Another interesting thing is, that wavelength of the pattern and even position of the peaks don't change if I shorten the stem. Other two lines represent the experiments where I cut the upper 10 and 20 cm of the stem and stimulated 1 cm below that.

The sketch of the stem you provided in your first post is correct. When the insect sings, it can obviously stand on any point, we don't know if they pick the position according to what their vibrating causes in the stem.

This part I don't understand. Are you saying that I should see/detect 10-cm long waves cycling at a rate 6 cycles per second? If that is so, then we're probably not talking about the same thing. Signal frequency is always approx. 100 Hz, and the pattern of amplitudes doesn't change with time (although I didn't try yet what happens if I stimulated with continuous 100 Hz tone).

Unfortunately no. I'll try asking at the relevant department of our university, but that may take a while, so let's keep this assumption for the time being if you feel it's good enough.

You're right about transverse waves being the most feasible for an insect to induce, but what about surface waves (Love, Rayleigh & co.)?

I see what you mean. Just to explain in case it matters: the insects we study don't produce vibrations by fluttering, but by the action of a special structure at the back of their abdomen - the tymbal. This is a hard membrane with a ridge in the middle which "clicks" when a corresponding muscle bends it and then "clicks" back to the same shape. Muscle activation is phase-locked to the frequency of vibrations - they contract 100 times per second. The resulting vibration then transmits through legs (where receptors are located) onto the surface. It's a fascinating system in many ways.

If your model is correct, the wind explanation is very much possible. As for the insects, they produce vibrations in the range of 70-130 Hz and this corresponds very well to what they react to.

Last edited: Jan 12, 2009
12. Jan 12, 2009

### skeleton

You need to determine the wave speed to calibrate the model. The speed is the same regardless of the frequency for the various modes. So this is how you determine speed:
- Uncouple the mini-shaker from the stem. Now the stem should be a true cantilever. I am hoping the stem remains upright as it is now unsupported at its top.
- Momentarily pull back the top of the stem and let it freely oscillate back and forth - like an inverted pendulum. This will vibrate now in its first mode. The wave length is known to be 4x the length of the stem (lambda=4*1.65m).
- Measure the period of oscillation, T1.
- The speed is calculated as: v = lambda/T1.

- You can then insert the wave speed into the mathcad sheet as I composed and see which vector element gives you a value close to your forcing function, f=100 Hz. If the corresponding wavelength matches your observed wave length (lambda.n=2*20cm=40 cm), then you have confirmed the mathematical model matches your test.
- If it doesn't match, then first scrutinize the assumed parameters, afterwards question the boundary conditions for representing the reality.
- My model assumed a one end fixed while the other end free - where the mini-shaker is. Alternatively, a different model might fit better; this would be one end fixed and the other end being a pinned hinge. For this case, the K vector is:
K = (15.4, 50.0, 104, 178, 272).

- Both mathematical models are for a beam of uniform density which alone contributes to accounts for inertial resistance (mass*acceleration). This is valid if the weight of the insect is insignificant, such that:
Mass(insect) < 1/5 * Mass(stem).
- Otherwise a different set of K values and corresponding equation would need to be used, to reflect the contribution of the insect's weight in offering inertial resistance to the forcing function.

Last edited: Jan 12, 2009
13. Jan 13, 2009

### skeleton

Your graph shows nodes spaced at 20 cm. Thus the wavelength is 40 cm. If you are uncclear then just sketch a sin wave transcribed thru both the positive and negative swings.

I think I can even see the 1st harmonic. This is prescribed indirectly by the interconnecting all the local maximums, despite this being a graph of velocities.

14. Jan 13, 2009

### yerpo

You mean like this?

This is getting more interesting by the minute. Although I still can't see the 40 cm wavelength - it's closer to 30 than 20, but still... Please correct me if I got it wrong.

I'm afraid the stem is too weak to stand upright by itself when the plant is this tall, but I'll try oscillating a shorter one. As for the insect, it weighs around 130 milligrams, which is pretty much negligible compared to stem weight.

15. Jan 13, 2009

### skeleton

Yes. For clarification, you can realize that the ends of a oscillating cantilever with sway will have a higher peak velocity then diminishing as you move closer to the fixed base. My green line is a better regression and consistent with theory.

I agree the wavelength (Lambda) is closer between 25 and 30 cm, not 40.

Just hang the stem upside down; it will still oscillate with the same periodicity. Just make sure the base (at the top) is fully fixed so it is a true cantilever. If not, the period will be wrong.

#### Attached Files:

• ###### velocity_pattern4.png
File size:
8.6 KB
Views:
154
Last edited: Jan 13, 2009
16. Jan 13, 2009

### yerpo

I'm not sure about your regression line. I left out point zero deliberately, because it's recorded directly opposite of the shaker, which is fixed and cannot oscillate freely, so it more or less registers its movement.

17. Oct 23, 2009

### yerpo

I'm resurrecting this thread because I am still not any closer to the solution, so please bear with me.

I tried what Skeleton proposed and determined the first harmonic. On a 83 cm tall stem, the period is almost exactly 1 second (around 980 ms, to be precise). But putting this value in the equation given above is problematic, because I don't know the speed of propagation. Waves in plants are dispersive, meaning that different frequencies have different velocities of propagation. Any ideas?

18. Oct 23, 2009

### Bob S

Here is an interesting quote from:
Directional hearing is also very important in many insects, and has a time resolution of a few microseconds. It appears to be based on arrival time of spiking neuron pulses in their central nervous system. Because neurons can only spike several (a few) times a second, insects are not very good at sensing direction of continuous sounds (or standing waves?).
Spiders will want to get direction from vibration (like from insects trapped in their web), so vibration impulses might be better than continuous sinusoidal vibration. Standing waves are actually simultaneous traveling waves in both directions, so contain no direction information.
Bob S

19. Nov 9, 2009

### yerpo

I'm familiar with the work of Cocroft's group, thanks. Their system is a bit different, though, because the little buggers they study communicate using higher frequencies that don't transmit as effectively as low frequencies that our buggers use. Plus, they're smaller.

The thing you said about simultaneous waves made me think, though. If I measured at two points within the same observed wavelength, I should get peaks of particle movement exactly at the same time. If the phenomenon observed isn't a standing wave, the peak would be positioned accordingly to how fast the vibration propagates. Is this a correct reasoning and is demonstrating this a good enough proof for the existence of standing waves?

We argue on other (neurophysiological) grounds that the very beginning of the signal (where the rebound signal doesn't arrive yet) is important for localization, the rest of the pulse only serving as a species-specific signal, so the lack of directional information might not be as problematic.

20. Nov 9, 2009

### Bob S

If you have a taut rope tied between two distant trees or posts, and give it a quick hit in the middle with a stick, you will see traveling waves (moving distortions) moving toward the trees, and reflected waves coming back. the forward and reflected waves pass each other without interference. Insects and spiders can determine the arrival times of the waves if there is no reflection or if they do not overlap. If the waves completely overlap, they are called standing waves (equal amplitude simultaneous waves in both directions), and the amplitude depends only on the location between the two posts.

This site discusses the difference between traveling and standing waves:
http://galileo.phys.virginia.edu/classes/152.mf1i.spring02/Waves.htm
See animation
http://openlearn.open.ac.uk/mod/resource/view.php?id=289477

Bob S

Last edited: Nov 9, 2009