How do standing waves produce sound?

[Helphelphelp]

Hi
i've been taught that standing waves do not transfer a net energy. but standing waves are formed in instruments, and we can hear them - so some sound energy/kinetic must be transferred to surrounding particles to reach our ears?
how does this work? please could you help me if my understanding has gone wrong?

 

[BvU]

Hello 3h,

Good question. I definitely don't blame you for not immediately picking up a subtle difference, and my compliments for not taking everything you hear () for granted !

A standing wave in its simplest form is propagating in one dimension and going back from one end to the other. Like a distortion from the equilibrium position (straight) of a string. This distortion propagates in both directions along the string and bounces back at the fixed ends. If you use a bow to introduce just the right frequency, you get a nice tone. Mechanical energy goes with the waves, but since amplitudes are same and directions opposite, no net energy transport in a direction along the string.

However, the string gives off a small part of the mechanical energy to let the air around it vibrate as well. compression-expansion: sound ! This, however happens in a direction perpendicular to the string. So the string does lose mechanical energy and the vibration dampens out.

And sure enough, in actual musical instruments there is yet another mechanism: the case of e.g. a violin is also picking up the mechanical vibration (mainly via the bridge) and passes this on to the air. Due to its size compared to the string itself, this is a lot more sound than what comes directly from the string. (cf speakers with/without soundboard).

If you do it in vacuum: no sound, so considerably less damping!

Does this help ?

--

 

[Helphelphelp]

brilliant - thank you so much!

 

[UncertaintyAjay]

Thanks, I was wondering the same thing myself. So that's why the wood that you make the soundboard out of influences the type of sound you get, the bridge and then the sounboard itself vibrate.

 

[sophiecentaur]

i've been taught that standing waves do not transfer a net energy.

I wonder where you were taught that bald statement. It doesn't quite put the right slant on the topic. Yes, it is true that some of the energy goes back and forth but standing waves are never complete; there are no absolute nulls - just minima. If no energy is lost from a standing wave then it will go on for ever and (in the case of an audio frequency) no one would hear it and the minima would be actual zeros. The thing about standing waves in musical instruments is that their resonant frequency (/frequencies) define the note; they don't need to 'store energy'. In the case of a continuously excited instrument (wind or violin), energy is being supplied to keep the standing wave going and energy is lost, via the air, and heard as a note. In a plucked / struck instrument, all the energy goes in at the start and it decays in time. Different instruments will have different depths of resonance.
In a different context, when radio frequency power is fed to a transmitting antenna, it is mostly radiated because of the matching network but some will be reflected and form a standing wave in the feeder line - that can be just a nuisance but, if there is a significant resonance (reflections and an unfortunate choice of feeder length), you can get unwanted high voltages on the feeder and damage the feeder and the transmitter.

 

[PWiz]

I wonder where you were taught that bald statement. It doesn't quite put the right slant on the topic. Yes, it is true that some of the energy goes back and forth but standing waves are never complete; there are no absolute nulls - just minima. If no energy is lost from a standing wave then it will go on for ever and (in the case of an audio frequency) no one would hear it and the minima would be actual zeros.

Exactly. One can arrive at this conclusion by simply applying the law of conservation of energy.

 

[tech99]

In the case of a continuously excited instrument (wind or violin), energy is being supplied to keep the standing wave going and energy is lost, via the air, and heard as a note. In a plucked / struck instrument, all the energy goes in at the start and it decays in time. Different instruments will have different depths of resonance.Interesting that in a case where power is supplied continuously, such as an organ pipe or antenna, when the generator starts, it slowly supplies energy to build the standing wave and which is then stored for the whole time note is played. Then, during sustained use, the generator supplies only the energy being radiated plus losses. And when the generator is switched off, the energy stored in the standing wave is then released and is slowly dissipated in radiation and in losses.

 

[tech99]

Sorry I messed this reply up just now.
Interesting that in a case where power is supplied continuously, such as an organ pipe or antenna, when the generator starts, it slowly supplies energy to build the standing wave and which is then stored for the whole time note is played. Then, during sustained use, the generator supplies only the energy being radiated plus losses. And when the generator is switched off, the energy stored in the standing wave is then released and is slowly dissipated in radiation and in losses.

 

[tech99]

Interesting that in a case where power is supplied continuously, such as an organ pipe or antenna, when the generator starts, it slowly supplies energy to build the standing wave and which is then stored for the whole time note is played. Then, during sustained use, the generator supplies only the energy being radiated plus losses. And when the generator is switched off, the energy stored in the standing wave is then released and is slowly dissipated in radiation and in losses.

And further to this, the energy for radiation and losses being supplied per cycle by the generator during continuous operation is much less, by the factor Q, than that stored in the standing wave . The radiation/loss energy is delivered by a pure traveling wave on the antenna or pipe, and is normally much smaller then the standing wave and is superimposed upon it. This is why the pattern does not quite become zero at the nodes. The residual voltage or current at a node is due to the traveling wave alone and is in quadrature with that of the standing wave elsewhere on the pipe.

 

[BvU]

Let's keep it simple. The green B with this thread isn't there for nothing!

 

[sophiecentaur]

Let's keep it simple. The green B with this thread isn't there for nothing!

I don't think it would be stretching things to include the fact that the width of the resonance response (to the range of frequencies it will respond to) is directly related to 1/Q. High Q means a narrow (sharp) response and well defined frequency and a long 'sustain' time.