# Intensity of a Loudspeaker, What's wrong with calculation?

1. Sep 5, 2011

### Malavin

1. The problem statement, all variables and given/known data
A certain loudspeaker system emits sound isotropically with a frequency of 1770 Hz and an intensity of 0.892 mW/m2 at a distance of 5.66 m. Assume that there are no reflections. (a) What is the intensity (in mW/m2) at 30.8 m? At 5.66 m, what are (b) the displacement amplitude and (c) the pressure amplitude of the sound? Take the speed of sound to be 343 m/s and the density of air to be 1.21 kg/m3.

I mostly need help with part b, I believe I can get part c with the answer from part b.

2. Relevant equations

I = 1/2(ρvω2sm2)

3. The attempt at a solution

So, for part b and from the relevant equation:

sm = (2I/ρvω2)1/2

ω = 2πf = 2*3.14159*1770 = 11121.24 Hz

So, using the above equation and plugging in I = 0.892 mW/m2, ρ = 1.21 kg/m2, and v = 343 m/s, I get the answer 5.895 x 10-6 m. However, this answer is apparently wrong, and I'm not sure why. Is there something wrong with my calculations?

Last edited: Sep 5, 2011
2. Sep 5, 2011

### Rayquesto

intensity is the power/area covered. Such areas are usually spherical. so at a distance of 5.66meters,

.892E6W=
8.92E5W/m^2=power(dE/dt)/4pie(5.66)^2 if this really is a sphere and get power using algebra.

once you get power, intesity=power/4pie(30.8)^2 I dont get neither b or c. lol I didnt study pressure yet. I don't even know what amplitude affects.