Question

Hi everyone, I have a problem about physics.

The question is as below:

A certain loudspeaker system emits sound isotropically with a frequency of 1740 Hz and an intensity of 0.853 mW/m² at a distance of 6.79 m. Assume that there are no reflections. (a) What is the intensity at 40.6 m? At 6.79 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.

The hint:

How does intensity depend on the power of the source and the distance from the source? Can you find the value of the power? Can you then find the intensity at the second distance? Can you relate the intensity there to the displacement amplitude? Can you relate the latter to the pressure amplitude?

Thanks a lot!

Answer 1

Hello,

As there are no reflections, the intensity of sound follows inverse square distance relationship.

i.e., I1/d²

Given that at distance 6.79m the intensity is 0.853 mW/m²; Let us assume intensity at 40.6m is x mW/m².

Then 0.853/x = 40.6²/6.79² ==> x=0.0238 mW/m².

The relation between intensity of sound and its displacement amplitude is given by

I = 2π²ρf²v∆s² (ref link:https://physics.info/intensity/)

substituting the values,

Solving for ∆s gives ==> (I/2π²ρf²v)^1/2 ==> (0.853*10^-3/2*3.14²*1.21*1740²*343)^1/2 ==>0.185m

The pressure amplitude ∆P and Intensity of sound are related as I = ∆P²/2ρv

Therefore solving for ∆P gives==>(2*0.853*1.21*343)^1/2 ==>26.6 Pa