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I've recently become fascinated with the idea of sound energy having a theoretical equivalent mass. I've read over this thread:

Do light and sound waves have mass

I understand this part: $m_{eq}=E/c^2$ and $E=A\rho \xi^2\omega^2$

Where I am getting tripped up is the way to measure $E$.

Essentially, I want to measure $E$ and eventually $m$ of a song (for it's entire duration). The only equipment I have is an iPhone app that measures intensity (dB) (apparently it's fairly accurate for levels below 100 dB). I plan to play the "music" at about 80dB over speakers in a room that is 50'x30'x15' and I will assume the temperature is at room temperature. If your wondering, it's for a conceptual audio art piece.

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Welcome to Physics.SE. We have the MathJax engine running on the site which allow us to write mathematics in a LaTeX alike language. There are some minimal notes on it in the FAQ. –  dmckee Mar 11 '13 at 21:49
    
If you like this question you may also enjoy reading this Phys.SE post. –  Qmechanic Mar 11 '13 at 22:13
    
Thanks dmckee. Qmechanic-an interesting indeed! Thx! –  user21863 Mar 12 '13 at 4:32

2 Answers 2

up vote 1 down vote accepted

A song has no mass equivalence. The sound waves from playing a song do---to the extent that they carry energy, and you can relate an equivalent mass1 to energy.

To cut to the chase, it would probably be easiest to look up how much power (energy per unit time) your speakers produce, then multiply that by the duration of the song.

If you measure the 'intensity' of the song with your iphone, that tells you the power-density at the location of the phone---while the speakers are actually generating sound-waves in all directions. Through an elaborate series of calculations and approximations, you could convert this to an energy. For completeness:
The number of decibels actually measures the pressure of the sound waves, which you can convert to an amplitude. You can estimate the frequency ($\omega$) as, something like, the higher frequency end of the human audible range (possibly adjusted for the type of song). The total power is then computed by finding the surface-area of a sphere around the source ($A$), at the distance you are measuring.

Again, using the power of the speakers will be more accurate.

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FWIW, people ask about measuring dB SPL on the iPhone. You would need to calibrate your iPhone first. Use a sound level meter instead. 50 bucks from radio shack. –  Bjorn Roche Mar 12 '13 at 3:32

You could also set an upper bound if you know the phone's battery capacity (usually written on the battery) and how long you get out of a full charge playing the song on repeat. You can estimate the energy required to play the song by

$$ \text{Energy per song} = \text{Battery capacity} \times \frac{\text{Song length}}{\text{Battery life}}$$

This is an upper limit because because, obviously, not all of the battery power goes into playing the song. In fact most will go into other things like wifi and heat. If you want a more accurate estimate you can turn off as many other things as possible. If waiting for the phone to die isn't doable then you could run down say 30% of the battery and correct for that, but discharge curves aren't necessarily linear.

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