# Relating Power output of a sound wave to the pressure...

1. Jul 22, 2015

### rwooduk

... it's amplitude exerts.

During cavitation a sound wave is applied to a liquid and breaks it apart and gas pockets are formed. The frequency and the amplitude of the sound wave effect the bubble/s.

My question is, if I have the power output of the device then how would this relate to the "pressure amplitude" or "drive-pressure amplitude" exerted on the bubble/s?

The pressure amplitude as the name suggests is usually given in atmospheres, I'm struggling to somehow get Watts to atmospheres.

For example a paper might say "the frequency and the amplitude of the ultrasound are 20KHz and 1.4 bar" how would I relate the 1.4 bar to the power output of the sound wave?

edit

Thinking back, power is proportional to the square of the amplitude (I am unsure of what the constants would be to make it equate, especially if the medium is water), so if I square root the power, but then how would I get to pressure?

Last edited: Jul 22, 2015
2. Jul 22, 2015

### rwooduk

hmm got it (below image), although I am unsure of what area squared I would divide power by to get intensity, so any ideas would be more than welcome.

3. Jul 23, 2015

### rwooduk

I'll update this.

It seems during experimental work a microphone transducer is used to measure the acoustic driving pressure, rather than it being calculated. Although I am unsure of how numerical simulations come up with a number and I have yet to find a paper that goes into detail on this (rather than just stating the pressure amplitudes).