Sound Waves Legacy Problem #7 Guided Solution
Problem*
The intensity of sound waves decreases as the distance from the source of sound increases. The relationship between intensity (I) and distance (d) is an inverse square relationship which follows the equation I = P/(4•π•R2) where P is the power of the sound source, usually expressed in Watts. Jake recently purchased a stereo system for his basement recreation room. Determine the maximum intensity of the sound waves at the following distances from his 120-Watt main speaker.
- 1.0 meter
- 2.0 meter
- 3.0 meter
Audio Guided Solution
Sources of sound put out sound waves, or sound energy, that travels through space in a three-dimensional fashion. The rate at which the sound energy is put out is referred to as the power of the source. It's a characteristic of the sound source, not so much the characteristic of the sound waves. Now these sound waves put out by the source will have an intensity in any given location from the source that varies inversely with the distance squared from the source. We refer to this as an inverse square law. We can calculate the intensity by taking the power and dividing it by 4 pi r squared, which happens to be the area of a sphere. And so, if I wish to find the intensity at a distance of one meter from a source having a 120 watt power rating, then I need to go 120 divided by 4 divided by pi divided by 1 squared. When I do that, I get 9.5493 watts per meter squared. That's the unit on intensity, and I can round that to two significant digits. I can repeat the calculation for two meters from the source. I would go 120 watts divided by 4 divided by pi divided by 2 squared, and this time I'm going to get 2.3873 watts per meter squared, which happens to be one-fourth of the answer from part A. That's the inverse square law. Double the distance, and you end up getting one-fourth the intensity. And then for part C, I can repeat the process and go 120 watts divided by 4 divided by pi divided by 3 meters squared. And that gives me 1.0610 watts per meter squared, which I can round to two significant digits. And when I'm done, I notice that in part C, the intensity is one-ninth of that of part A, which again is the inverse square law, that if you triple the distance, you get an intensity value that is one-ninth, or 1 divided by 3 squared, of the original intensity.
Solution
- 9.5 W/m2
- 2.4 W/m2
- 1.1 W/m2
Habbits of an Effective Problem Solver
- Read the problem carefully and develop a mental picture of the physical situation. If necessary, sketch a simple diagram of the physical situation to help you visualize it.
- Identify the known and unknown quantities and record in an organized manner, often times they can be recorded on the diagram itself. Equate given values to the symbols used to represent the corresponding quantity (e.g., \(\descriptive{v}{v,velocity} = 345\unit{\meter\per\second}\), \(\descriptive{λ}{λ,wavelength} = 1.28 \unit{m}\), \(\descriptive{f}{f,frequency} = \colorbox{gray}{Unknown}\)).
- Use physics formulas and conceptual reasoning to plot a strategy for solving for the unknown quantity.
- Identify the appropriate formula(s) to use.
- Perform substitutions and algebraic manipulations in order to solve for the unknown quantity.
Read About It!
Get more information on the topic of Sound Waves at The Physics Classroom Tutorial.