Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Lost Energy in Sound Waves?

  1. Dec 29, 2004 #1
    This is a simple question, but i can't seem to get my bearings right on it. Consider two speakers at essentially the same point emitting sound waves that are half a cycle out of phase but otherwise identical. Interference dictates that these two waves should cancel out. The resultant wave, now "flat", appears to deliver no energy as it "travels" outward. But this appears the contradict that each speaker is delivering energy to the air as it generates its respective sound wave. Could someone give me a solid understanding of what is really going on? Thanks.
     
  2. jcsd
  3. Dec 29, 2004 #2

    StatusX

    User Avatar
    Homework Helper

    If the sources aren't at exactly the same place, then in a certain direction they will exactly cancel, but in a perpendicular direction they will constructively interfere, and the total energy output would be the sum of that from the individuals.
     
  4. Dec 29, 2004 #3
    I'm not so sure I would agree. Suppose the speakers are seperated by distance much smaller than the wavelength of the emitted waves. Moreover, the each wave is one half-cycle out of phase. Then there is destructive interference along points equidistant from each speaker, but no constructive interference.
     
    Last edited: Dec 29, 2004
  5. Dec 29, 2004 #4

    StatusX

    User Avatar
    Homework Helper

    You're right, and I realized this a little while ago but wasn't sure how to fix this problem until now.

    First note that acceleration is half a cycle out of phase with displacement in a sine wave. Start with a single emitter, and start ouptutting a steady sine wave signal. Now, what you must remember is that these emitters are physical objects, and they will respond to the medium as easily as they affect it. So when you place the second emitter close to the first, without supplying it any external power, it will be affected by the waves from the first and start moving a half cycle out of phase.

    You can either interpret this as that the first emitter is using most of its power to move the second, or they are both emitting waves that destructively interfere. The power applied is equal to the power delivered, because much of this power is used to move the second emitter. Another way of thinking of this is that the power required to get an emitter to output a wave of a certain amplitude depends on the current excitation in the medium, and this value will come out to be negative in this case.
     
    Last edited: Dec 29, 2004
  6. Dec 29, 2004 #5
    Let me get this straight. Suppose both speakers are affixed to the ground and close to each other. Furthermore, suppose the speakers are one half-cycle out of phase so that when the first attempts to compress the nearby air, the second is expanding this air at the same rate. Since these speakers are otherwise identical, the pressure of this nearby air will then remain constant. The average power delivered by each speaker then falls to zero because the work done on the air by compressing it will be exactly cancelled by the work the air does on the speaker during its expansion. I think that's what you were getting at, that each speaker affects the motion of the other.

    I'm still not sure if this is a sound explanation.
     
  7. Dec 30, 2004 #6
    genxhis,

    can you be more specific as to what you are doing? if you are talking about MUSIC being played, and trying to "tune" the placement of your speakers, that is one thing. if you are just making an experiment, and using monotone frequencies, that is another.

    TRoc
     
  8. Dec 30, 2004 #7

    krab

    User Avatar
    Science Advisor

    Yes. The emitted sound power drops to zero. But so does the speakers' impedance. IOW, there are very large cone excursions for very small power. I have some experience with this. I had a car with a totally enclosed trunk. I put two speakers on the back dash and initially had them out of phase. The result was that I had no bass, but nevertheless the speakers were very easy to overdrive; I nearly blew them up before I found the problem.
     
  9. Dec 30, 2004 #8
    T. Roc. Oh, no. I am not "really" trying to design experiments. I am just trying to make sense of conservation of energy and the principles of interference together. For instance, two identical speakers near each other and out of phase will cancel out any waves. My question was then "where does the energy output by the speakers go?".

    Krab. It's interesting that you've actually seen this in action. In fact, I was thinking that the speakers might experiance growing amplitudes. As you said, the speakers' drivers will no longer have to overcome resistances. And where else would the energy be invested?

    But suppose the speakers are far enough from each other that they do not interfere with the other's operation. Is it possible to show formally that the power delivered by the resultant wave is the sum of the power delivered by the individual waves? This question stems from the following remark in my text: "The intensity due to a number of waves is the sum of the individual intensities." Note that "intensity" here is power per unit area. Or was this only meant to apply when there isn't strong interference?
     
  10. Dec 30, 2004 #9

    StatusX

    User Avatar
    Homework Helper

    With one source interfereing with itself by, say, two slit diffraction, there will be constructive and destrucitve interference at different points, and the total power at all points will add to the power put out by the emitter. With two or more emitters interfering, the situation is a little more complicated because the emitters must not only provide enough power for their own waves, but also enough to compensate for the waves from the other emitters that are affecting them. In general, if there is constructive interference, this extra necessary power will be positive, and if there is destructive interference, it will be negative. By negative power, I mean instead of doing work on the medium, the medium does work on it. So if you are using electric speakers in the original situtation, when you first hook up the second one, it will act as a microphone and actually deliver extra power into the circuit.

    I realize in my last post I treated the speakers very assymetrically, and this was inaccurate. In reality, they will reach an equilibrium where they each help drive eachother and very little external power is needed to keep them in this motion. If you continue providing the normal power as if there were no other speaker present, what you are doing is driving a simple harmonic oscillator at resonance, and it will reach very high amplitudes very quicky. In fact, if it doesn't blow out first, these amplitudes will get so high that the tiny waves that go off to the side and aren't completely interfering will be strong enough to carry all the power you are putting into the system.

    If the speakers are farther apart, there will be less power delivered to each by the other speaker, but there will also be less destructive interference, so energy is still conserved.
     
    Last edited: Dec 30, 2004
  11. Dec 30, 2004 #10
    It's almost like it turns into a balanced force. Interesting question...
     
  12. Dec 30, 2004 #11
    genxhis,

    krab has pointed out another difference - when 2 speakers share the same "-" air space, which is different than 2 separate speakers sharing "+" air space (and not "-" space, ie. the enclosure). by negative, i mean the downward movement of the cone, and positive = upward movement. the - movement creates compression in the + space, and simultaneously, expansion of the - space (and vice versa).

    we know that amplitude (energy) does not change the frequency, but a change in the medium (density) can change the wavelength. krab said "i had no bass", which, although he is at least double the physicist that i am, i would say is not correct from a physics perspective. i think his wavelengths increased to the point below the threshold for human hearing. with the speakers being out of phase, and given the strength of the enclosure (metal trunk), one speaker is always "enticing" this extra wave-length through the weakest point - the cone material, which causes distortion of the original signal.

    this is one reason that "seekers of extreme bass" use separate bridged amplifiers for bass, preferring the consistency of mono to the quality of stereo. until you are in the higher frequencies, stereo is not needed. a band pass filters these frequencies in the most efficient way. mono eliminates "out of phase" signals.

    the lower frequencies suck up the most energy because they move the most mass (through the longest "throw" of the cone). this is where you will find the conservation of energy (or its' equivalent - mass).

    when you mentioned having the waves cancel each other out, remember that your position is part of the equation. if you are equidistant from each speaker that is out of phase, it "cancels out", but if you move, this effect disappears. this demonstrates that most of what we are talking about is PERCEPTION, and not the math behind it (physics)- it is not as difficult as it might seem.

    TRoc
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Lost Energy in Sound Waves?
  1. Energy in sound waves? (Replies: 8)

  2. Waves of Sound ! (Replies: 8)

  3. Sound waves (Replies: 11)

Loading...