# Sound intensity/Ear drum problem

1. Feb 25, 2015

### Zeus711

1. The problem statement, all variables and given/known data
A source emits sound waves equally in all directions. The intensity of the waves 2.50m from the source is 1.9 * 10^-4 W/m^2
a) Find the power of the source
b) If the diameter of your ear drum is 8.4 mm, how far from the source do you have to be located, so that your ears combine receive 0.42*10^-12 Joules of energy.

2. Relevant equations
Intensity = Psource/(4πR^2) Due to it being an isotropic source

3. The attempt at a solution
So part (a) I'm easily able to solve using the formula, Intensity = Psource/(4πR^2)
Just plug in the value for the intensity and distance (2.50m), isolate for Psource, and you get .015 Watts.

Part (b) I'm pretty confident I know how to solve, there are just a few things that are tripping me up. What I would do is
1) Energy = (Preceived) * (time) From this you are able to determine the Preceived is 0.42*10^-12 Watts, by simply plugging in 1 second for time.
2) You would then use Preceived = (Intensity)/(Area). Using the diameter of the ear given, we are able to solve for the area: (π(4.2*10^-3 m)^2). Plug in the value from Preceived and get the Intensity.
3) Plug this Intensity value, along with Psource from part (a) into the formula, Intensity = Psource/(4πR^2)
and then solve for R. Am I right in determining that the 'R' in this formula would represent the distance we are trying to determine?

Can someone confirm this process is correct?
The issue that I am running into is that the question asks how far you have to be so that BOTH your ears combined receive 0.42*10^-12 Joules of energy. From our calculations, we only took into account the area for one ear. Do I double the area for the ear for the prior step then solve? Or can I just halve the power and then solve for one ear? I'm so confused.

2. Feb 25, 2015

### haruspex

Where do you get the period of 1 second from? Seems to me the question is flawed. It should either specify a period of time or specify a level of received power, not a quantity of energy.
Other than that, yes, just double the area.

3. Feb 25, 2015

### Zeus711

Oh yeah. Forgot to add that the question says 0.42*10^-12 Joules of energy per second.
Would my calculations now be correct?

4. Feb 25, 2015

Should be.