Frequency in oscillating membrane

In summary: Well, I'll do some experimenting with the nodes and see how long I can make the reverb. Thanks for the tip.In summary, the conversation discusses building a plate reverb and the factors that affect its frequency response. The set-up includes a rectangular plate with forced motion and multiple pickups. The main concern is how the plate's thickness affects the frequency response. The conversation also delves into the wave equation and harmonic oscillation to calculate the frequency response, as well as the placement of nodes for optimal reverb effect. The conversation also touches on the importance of the plate's size and tension for achieving the desired sound, as well as the use of acoustic damping material for controlling the reverb time.
  • #1
liquidFuzz
97
3
I'm about to build a plate reverb and I have some thought, or questions about the plate. The set-up is as follows:

A rectangular plate, firmly tensioned in the corners.
A forced motion entered.
2 or more pick-ups, piezo style.

I'd like to know how the plate, it would be thickness, affects the frequency response in the plate.

[tex]\rho \frac{\partial^2u}{\partial t^2} = S\nabla ^2 u[/tex]
Where [tex]\rho[/tex] = desity; mass/area
S = tension; force/length
u = amplited; length from idle.

Am I poking around in the right mud pit..?
 
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  • #2
Hmm, no real action in this forum..? I did some calculations on a rectangle with length 2a and width a. Tension/density is set to c^2, initially to easy down the bloatyness

[tex]\nabla^2u(t,x,y) = \frac{\partial^2u(t,x,y)}{\partial x^2} + \frac{\partial^2u(t,x,y)}{\partial y^2}[/tex]

u(0) = on the edges. Not the real case as I'm planing to use one fixation on the short side and two on the length. (All in ideal nodes, half, thirds.)

Wave equation with c as above:
[tex]\ddot{T}~X~Y = c^2 T \frac{d^2}{dx^2}~X~Y + c^2 T \frac{d^2}{dy^2}~X~Y[/tex]
Harmonic oscillation:
[tex]\ddot{T} = -\omega^2 T [/tex]

Then
[tex]\frac{\frac{d^2x}{dx^2}}{X} = - \frac{\frac{d^2y}{dy^2}}{Y} - \frac{\omega^2}{c^2}[/tex]

Further set lambda
[tex]\frac{\frac{d^2x}{dx^2}}{X} = -\lambda^2[/tex]

With x(0) and x(2a)

Try a solution
[tex]x = A sin(\lambda x) + B cos(\lambda x)[/tex] with [tex]x(0) = 0 \Rightarrow B = 0[/tex][tex]x(2a) = 0 \Rightarrow sin(2a\lambda) = 0[/tex]Then [tex]\lambda = \frac{n\pi}{2a}, n = 0,1...[/tex]

Now [tex]\frac{d^2Y}{dy^2} + \left[ \frac{\omega^2}{c^2}-\left( \frac{n\pi}{2a}\right)^2\right] = 0[/tex]

Same solution for y.
[tex]Y= D sin\left(y\sqrt{\frac{\lambda^2}{c^2}- \left(\frac{n\pi}{2a}\right)^2}\right) + E cos\left(y\sqrt{same...}\right)[/tex] with [tex]y(0) = 0 \Rightarrow E = 0[/tex][tex]y(a) = 0 \Rightarrow sin\left(a\sqrt{\frac{\lambda^2}{c^2}-\left(\frac{n\pi}{2a}\right)}\right) = 0[/tex]

Then
[tex]a \sqrt{\frac{\lambda^2}{c^2}-\left(\frac{n\pi}{2a}\right)^2} = m\pi, m=0,1...[/tex]

[tex]\omega_{n,m} = \frac{c\pi}{a}\sqrt{\frac{n^2}{2^2}+m^2}[/tex]

The nodes I'm aiming for gives n = 3 and m = 2 gives and c, let's throw in a = 1 meter while we're at it.
[tex]\omega_{3,2} = \sqrt{\frac{S}{d}}\frac{\pi}{1}\sqrt{\frac{3^2}{3^2}+2^2}[/tex]

Finally
[tex]\omega_{3,2} = \sqrt{ \frac{S}{d} \pi^2 5}[/tex]

I tried to tinker with a induced force but I could remember how to do it... Anyone care to pitch in..?
 
  • #3
The tricky part seems to be your boundary conditions. I've found solutions for a rectangular plate with free edges:

http://www.sciencedirect.com/science/article/pii/S0022460X10004001

elastically restrained edges:

http://qjmam.oxfordjournals.org/content/12/1/29.abstract

And a variety of fixed or free edges configurations:

http://archive.pepublishing.com/content/c83n7p76676r8584/

But nothing for fixed corners. The third reference may have this particular configuration, but I didn't go through it in detail.
 
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  • #4
One of the tricky things with a plate reverb is to get a good signal at high frequencys. I just figure I'd get more high pitched tones if I forced the membrane into overtones. That's why I'll try placing the nodes in a half by third manner. Though, every plate reverb I see has the nodes in the corners... Any thoughts on that?

(It's not that hard to move the fixation/nodes. Just a bit of tinkering...)
 
  • #5
I'm not an expert about this technology, but it seems that the high end response of the sheet will be primarily dictated by how thin and the specific material the sheet is made of- possibly, a very thin, stiff, low density material is brighter- the tensioning will control the low and midrange tone response.

http://www.prosoundweb.com/article/how_to_build_your_own_plate_reverb/
http://www.beavisaudio.com/Projects/Plate_Reverb/index.htm
http://nicksworldofsynthesizers.com/plate.php
 
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  • #6
You're quite right. My idea isn't that good. A smaller plate would have virtually the same effect as having multiple nodes.
 
  • #7
You are probably worrying about the wrong thing in trying to calculate the mode frequencies.

For a good reverb effect, you want to be in the frequency range where there are a large number of modes with closely spaced frequencies. Otherwise, you will hear the "ringing" of individual modes coloring the sound, like singing in a bathroom. So the important number is the lowest mode frequency, which you want as low as possible. That means the plate should be as big and thin as possible (something like 1m x 2m x 0.5mm thick steel).

When you tension the plate by pulling on the corners, the stress distribution in the plate is very non-uniform. This is a good thing, because it means the wave speed varies across the plate, so the sound travels through the plate and is reflected off the edges in a complex fashion compared with a simple system with a small number of resonances at different frequencies, but it means it is hard to estimate the lowest resonant frequency. In practice, you change the tension till yoo get the sound you want, rather than trying to calculate anything. You can easily "measure" the lowest frequency by tapping the plate and listening.

You probably want to filter out the low frequencies in the audio that you send to the reverb, to avoid exciting the lowest frequency modes of the plate.

Another thing you might want to consider is the damping factor (how long the reverb will ring for). The traditional way to control that is to use another plate covered with acoustic absorbing material, close to but not touching the reverb plate. Changing the air gap between the plates will change the reverb time (smaller gap = less reverb).
 
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  • #8
Zero - Lol, I'm building tube-amps and I like to calculate frequency, voltages and **** in them. All my friends keep telling me to drop it, 'just build the freaking amp'. You took me by surprise I must say. Any other physics nerd telling me to easy down. :-)
 
  • #9
Anyhow, I'm very grateful, thanks for al the input. I'll let you know how it turns out.

(The hardest thing seem to be gathering all the right stuff. Like the plate it self...)
 
  • #10
liquidFuzz said:
Zero - Lol, I'm building tube-amps and I like to calculate frequency, voltages and **** in them. All my friends keep telling me to drop it, 'just build the freaking amp'. You took me by surprise I must say. Any other physics nerd telling me to easy down. :-)

Don't get me wrong, I'm 100% in favour of doing useful calculations as part of the design process.

But trying to calculate all the modal frequencies of a pre-stressed plate is
(1) hard, unless you make a nonlinear finite element model of it
(2) not very useful, because the results will be very sensitive to the exact way you restrain the plate and the stresses caused by the tensioning.

There are basically two ways to understand the response of a system like a plate reverb. One is to (attempt to) calculate the modal frequencies. A better way is to think about how the sound waves travel around the plate. Suppose you apply a very short pulse at your input transducer position. That pulse will spread out across the plate like ripples from a stone thrown into a pond, reflect off the edges etc, travel at different speeds depending on the amount of tension in the plate (possibly different speeds in different directions, and in different parts of the plate). Every so often, the pulse willl pass the output transducer and generate a pulse of output.

What you want to achieve is large number of output pulses (i.e. lots of reflections around the plate), with as "random" a distribution of times between them as possible. The exact details of the "randomness" don't matter much, so long as it really is random. For example, what you DON'T want is for the wave to travel back and forth along the length of the plate in a straight line, and produce a regular series of output pulses and nothing else.

Thinking about it that way explains why you don't want any symmetry in the positions of the input and output transducers, (i.e. don't put them at simple fractions like 1/2 or 1/3 the distance from the edges or corners, and certainly don't put the input and output at "mirror image" locations relative to each other.

It should be physically "obvious" that the two ways of analyzing what is going on must be related to each other, because they are two descriptions of the same thing. Indeed they are related mathematically, the first being a modal model or a frequency-domain model, the second being an impulse-response model. Any text on modal analysis will explain how the math of the different models is related.

All this also applies to amps etc. You can calculate and measure the frequency response, or look at the transient (or impulse) response, e.g. input a low frequency square wave, and use an oscilloscpoe to see what the amp does to it.
 
  • #11
Usful as in getting an idea of what the BEEP it is I'm doing..? ;-) Even if the result in this case was intuitive, like the symmetry your talking about.

As we're getting more pragmatic I'll let you in on my plans. The plan is to use double-sided tape so I can try different locations, this goes for the piezo pick-ups. The inducer however, I haven't come up with a way to mount it so that I can test different locations. So far I'm thinking like this. A rod crossing the plate sideways, shortest way, placed slightly off center. The inducer must now be mounted to the rod in a way that'll let me try different locations.
 
  • #12
Maybe an update of what I've got now.
[tex]\omega_{1,1} = \sqrt{\frac{S}{d}}\frac{\pi}{1}\sqrt{\frac{1^2}{2^2}+1^2}[/tex]

Finally
[tex]\omega_{1,1} = \frac{\pi}{2}\sqrt{ \frac{5S}{d} }[/tex]
 

Related to Frequency in oscillating membrane

What is frequency in oscillating membrane?

Frequency in oscillating membrane refers to the rate at which the membrane oscillates or vibrates. It is measured in Hertz (Hz), which represents the number of cycles or vibrations per second.

How is frequency related to wavelength in oscillating membrane?

The frequency of an oscillating membrane is inversely proportional to its wavelength. This means that as the frequency increases, the wavelength decreases, and vice versa.

What factors affect the frequency of an oscillating membrane?

The frequency of an oscillating membrane is influenced by several factors, including the tension of the membrane, its mass, and the density of the medium it is vibrating in.

How does the frequency of an oscillating membrane affect the sound it produces?

The frequency of an oscillating membrane determines the pitch of the sound it produces. A higher frequency results in a higher pitch, while a lower frequency corresponds to a lower pitch.

Why is understanding frequency important in studying oscillating membranes?

Frequency is a crucial concept in understanding the behavior of oscillating membranes. It helps us predict the motion and characteristics of the membrane, and is essential in various fields such as acoustics, music, and engineering.

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