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What's so special about resonant Frequencies?

by masscal
Tags: frequencies, resonant, special
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masscal
#1
May7-14, 05:00 AM
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What's so special about resonant Frequencies ( or length of the pipe) in an pipe:

As I understand it, if any two waves pass by each other in different directions, they will create a sort of standing wave with anti-nodes and nodes ( hence beats).

They always show diagrams of Displacement with an displacement antinode at the end (ex http://hyperphysics.phy-astr.gsu.edu...es/opecol.html ) . Why is that? What if the pipe wasn't a perfect length to create an anti-node displacement at the end. Wouldn't there still be nodes and anti-nodes in the pipe ? ( assuming it's long enough) . There would still be a reflected and incident wave interacting and creating nodes and anti nodes. So why does a particular frequency matter
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Simon Bridge
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May7-14, 08:11 AM
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Have you checked the definition of "resonant frequency" online yet?
This will tell you what is special about it.

But you seem to be talking about harmonic frequencies.

For a pipe open at both ends, there is no length that will not have an antinode at each end.
If the pipe is closed at one end, then there will be an antinode at the open end and a node at the closed end.
You can understand why this has to be when you consider carefully what the wave on the diagrams actually represents.

These are the sorts of things you can figure out for yourself - please attempt this.
Your attempt will help me to help you.
dauto
#3
May7-14, 08:53 AM
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If the pipe had a different length than the standing wave would have a different wavelength to fit the pipe. That's why different instruments with different length produce sounds with different frequencies.

masscal
#4
May7-14, 02:39 PM
P: 28
What's so special about resonant Frequencies?

Quote Quote by Simon Bridge View Post
Have you checked the definition of "resonant frequency" online yet?
This will tell you what is special about it.

But you seem to be talking about harmonic frequencies.

For a pipe open at both ends, there is no length that will not have an antinode at each end.
If the pipe is closed at one end, then there will be an antinode at the open end and a node at the closed end.
You can understand why this has to be when you consider carefully what the wave on the diagrams actually represents.

These are the sorts of things you can figure out for yourself - please attempt this.
Your attempt will help me to help you.

Why, because the air is free to move at the open end, and fixed at the closed end. ) .

But why does it matter which Frequency you feed into the Pipe. Shouldn't any Sound Frequency go through pipe Reflect off the other side, and then the Incident Wave and Reflected wave should create some sort of Superposition wave pattern with anti-nodes and nodes.
Nugatory
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May7-14, 02:49 PM
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Quote Quote by masscal View Post
Shouldn't any Sound Frequency go through pipe Reflect off the other side, and then the Incident Wave and Reflected wave should create some sort of Superposition wave pattern with anti-nodes and nodes.
The wavelength is related to the frequency, and the nodes have to separated by an integral number of wavelengths. Thus, only certain frequencies (those for which the length of the tube is an integral or half-integral multiple of the wavelength) can be superimposed to yield a standing wave pattern with a node or anti-node at each end of the tube.
masscal
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May7-14, 03:57 PM
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Quote Quote by Nugatory View Post
The wavelength is related to the frequency, and the nodes have to separated by an integral number of wavelengths. Thus, only certain frequencies (those for which the length of the tube is an integral or half-integral multiple of the wavelength) can be superimposed to yield a standing wave pattern with a node or anti-node at each end of the tube.
But what if it wasn't wasn't an integral multiple of the wavelength. There would still be Nodes and anti-nodes in the pipe, right?
AlephZero
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May7-14, 04:24 PM
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The point you are missing is that if there is an antinode at an open end of the pipe, almost all the energy in the traveling wave gets reflected back into the pipe, as a wave traveling in the opposite direction. The resonance builds up because of repeated reflections.

If there was no antinode at the end, most of the energy in the traveling wave would leave the pipe and spread out into the air.

If you put a loudspeaker inside the pipe, you can generate sound at any frequency you want, but you only get a resonance when most of the energy can't leave the pipe because there is an antinode at the end.
Nugatory
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May7-14, 04:31 PM
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Quote Quote by masscal View Post
But what if it wasn't wasn't an integral multiple of the wavelength. There would still be Nodes and anti-nodes in the pipe, right?
The velocity of the air has to be zero at the closed end of the pipe (because it can't move there) and the pressure of the air has to equal to the outside pressure at the open end of the pipe (because if it weren't, air would flow until until the pressures were equal).

Thus, those points have to be nodes or antinodes. But all the other nodes and anti-nodes stay put (this is a standing wave) only because they're exactly balanced by their adjacent anti-nodes and nodes, spaced exacly one wavelength apart. Thus, if have a node at one end of the pipe, the next node must be an even wavelength up the pipe, and so on until you get to the other end - and there has to be a node there to anchor the end of the chain.

It may be easier to visualize standing waves in a rope tied to a tree. You hold one end of the rope and wave your hand back and forth, and you can set up a standing wave in the rope. You'll always have a node (zero movement of the rope) at the point where the rope is tied to the tree analogous to the closed end of the tube, and an anti-node (maximum movement of the rope) at your hand, analogous to the open end of the tube. There's only room for an integral number, plus one-half, wavelengths in between, and you'll find that the standing wave pattern only forms when you move wrist at a frequency that produces one of those wavelengths.
masscal
#9
May7-14, 05:01 PM
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Quote Quote by AlephZero View Post
The point you are missing is that if there is an antinode at an open end of the pipe, almost all the energy in the traveling wave gets reflected back into the pipe, as a wave traveling in the opposite direction. The resonance builds up because of repeated reflections.

If there was no antinode at the end, most of the energy in the traveling wave would leave the pipe and spread out into the air.

If you put a loudspeaker inside the pipe, you can generate sound at any frequency you want, but you only get a resonance when most of the energy can't leave the pipe because there is an antinode at the end.
Does that mean that the energy from the sound keeps building up as it enters the pipe?
AlephZero
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May7-14, 06:50 PM
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Quote Quote by masscal View Post
Does that mean that the energy from the sound keeps building up as it enters the pipe?
Yes, that's what "resonance" means.

In real life, the antinode at the end of the pipe doesn't reflect the sound perfectly, and a small amount gets out (which is what you hear as the sound made by the pipe). Energy is lost from the air inside the pipe on other ways, for example because of the viscosity of the air. So the sound inside the pipe only builds up until all the energy losses balance the amount energy that is being supplied.
Simon Bridge
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May7-14, 08:57 PM
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Quote Quote by masscal View Post
Why, because the air is free to move at the open end, and fixed at the closed end. ) .
That's the idea - if the wave shape represents particle motion, then there is maximum available motion at an open end. It's like a string that is not fastened at the end.

I'd like to add to what the others have said for the next bit:
But why does it matter which Frequency you feed into the Pipe. Shouldn't any Sound Frequency go through pipe Reflect off the other side, and then the Incident Wave and Reflected wave should create some sort of Superposition wave pattern with anti-nodes and nodes.
That is correct - except that the wave reflects multiple times, with a mixture of constructive and destructive interference along the whole length of the pipe.

The dominant effect of reflections after the first is destructive interference - for off-resonance frequencies. At any time there is always some part of the wave destructively interfering with another part - at any position. Resulting in nothing present.

IRL. the losses mean that the volume of the noise decreases quickly as the frequency moves off-resonance.

This is why you can make a rude noise down a trumpet and get a nice note out... the nasty frequencies all interfere themselves out of existence.

You can see this happen if you work out what all the reflected waves will look like, sketch them one above the other, and add them up geometrically.


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