Can we somehow transform the noise wave sounds, from a highway, to electrical energy?

  • $\begingroup$ This is very interesting in my opinion. I think you can consider extracting power from noisy streets. While a transducer may be expensive, I believe the energy you get may be helpful in the long run. $\endgroup$ Jun 3, 2017 at 15:35
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    $\begingroup$ You're not talking about a microphone? If you're talking about generating enough power to replace a power plant, then there's a case to be made for tidal power to be viewed as generated by extremely low frequency sound. $\endgroup$ Jun 3, 2017 at 18:29
  • $\begingroup$ Mind that a really huge microphone would probably create aerodynamic conditions that would actually cause ever so slightly more fuel to be used :) We've all had bad nights with that "conservation of energy" b**ch, she's gonna trick us again... $\endgroup$ Jun 3, 2017 at 20:23
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    $\begingroup$ From memory, the sound power of a 12 person brass band is under 5 Watts, and a symphony orchaestra is 12-15 watts. So while its possible there's not enough energy available to make it useful. Speakers are only 1-2% effective at converting electricity to sound. $\endgroup$
    – Criggie
    Jun 4, 2017 at 1:04
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    $\begingroup$ I'm half inclined to post "Yes, you've just described how your own hearing works" as an answer... $\endgroup$
    – zzzzBov
    Jun 4, 2017 at 3:47

2 Answers 2


You could do this, but the amount of energy is too low to be useful.

Even if the noise from the highway was continuous and loud enough to cause long term hearing damage, the sound energy level would only be about 1 milliwatt per square meter. You would need several square meters of sensors even to power a single LED light, which isn't very practical.

Human hearing works on a logarithmic scale of loudness, not a linear one - a typical "noisy crowd of people" environment only has an energy of about 1 microwatt per square meter, and the threshold of human hearing is about 1 picowatt ($10^{-12}$ watts) per square meter.

See http://www.physnet.org/modules/pdf_modules/m203.pdf.

  • $\begingroup$ To me, the ability to transmit and record sounds is extremely useful. Are we assuming this question is about electric power like that which is generated by power plants? $\endgroup$ Jun 3, 2017 at 18:31
  • $\begingroup$ @ToddWilcox I was assuming it was about harvesting electrical energy. Recording and transmitting sound are obviously useful, but any practical recording or transmission system needs an external source of electrical energy - it doesn't work using only the sound energy captured by the microphone or other type of transducer. $\endgroup$
    – alephzero
    Jun 4, 2017 at 1:05
  • $\begingroup$ @ToddWilcox In other words, it's not converting the sound into electrical energy. The energy comes from another power source, and it's simply modulated to match the sound waves. $\endgroup$
    – Barmar
    Jun 4, 2017 at 5:34
  • $\begingroup$ This answer could be improved if it actually said how to do this, instead of just saying "You could do this". Sure, a simple "yes" is a valid answer, but it's not nearly as useful as it could be. $\endgroup$
    – user27542
    Jun 4, 2017 at 8:24

You are doing it whenever you speak in a mic . There is a transducer in the mic that is converting your sound energy to the electrical energy which is then converted back to the sound energy through speakers. In your idea we will have to use a very big or a very high tech transducer which may costs a lot and is not economical.

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    $\begingroup$ A single big transducer would not do the trick – a bigger diphragm becomes ever less sensitive to high frequencies; at some point you could at best utilise seismic waves, but not traffic noise anymore. Thousands of small mics could be used, but that's utterly uneconomical. $\endgroup$ Jun 3, 2017 at 17:41
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    $\begingroup$ "the electrical energy which is then converted back to the sound energy through speakers." Not converted back to sound energy. The signal is amplified, and that's what the speaker converts to sound. What I'm trying to point out is that the conversion is nowhere near as efficient as your explanation makes it seem: Losses are enormous and amplification is necessary. You don't really get the electrical energy into the speakers without adding lots of... Electrical energy, to the system. $\endgroup$
    – Beanluc
    Jun 3, 2017 at 18:11

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