You are listening to your stereo at a location where the intensity of the sound wave is 0.00014 W/m^{2}. What is the intensity of the sound wave if you move twice as far away from the speakers? (Hint: Treat the sound wave as a spherical wave.)
In this problem, you are given the intensity of a wave at one distance and asked the intensity of the same wave at another distance. Intensity of a wave is defined as the power per unit area, and area depends on distance from the source, so intensity is closely related to power through distance. This is a definition of intensity problem.
You might notice at this point that you don’t have enough information (you don’t know distance) to plug directly into the definition equation. If you don’t notice that now, don’t worry about it. It will become apparent in the solution. However, if you do notice it now, it should not stop you from continuing with the solution. A very common approach when you are given information about one point and asked for similar information about another is to take the ratio of the definition equation at the two points. In this way, the unknown information is divided out.
This definition problem will be solved taking ratios.
There is no need for a picture in most definition problems, and this is one of them. You are given distance and intensity information at one point and asked for the intensity at another distance. A picture will not provide any additional organization beyond what is already present in the problem. However, if picturing the wave helps to understand the equation, then you should absolutely make any sort of sketch that is helpful to you.
In equation form, the intensity of a spherical wave is given by
I = Power / Area = P / (4πr^{2})
This is the only relation you need for this problem.
I = P / (4πr^{2})
If you didn’t recognize it earlier, you now see that you cannot answer the question just by plugging in information about the point of interest (your new location.) But you are given information about two different points, and so you can solve the problem by taking the ratio of this equation at the two points:
I_{2} / I_{1} = [P / (4πr^{2})]_{2} / [P / (4πr^{2})]_{1}
I_{2} / I_{1} = P(4πr_{1}^{2}) / P(4πr_{2}^{2})
I_{2} / I_{1} = r_{1}2 / (2 r_{1})^{2}
I_{2} / I_{1} = 1 / 2^{2} = 1 / 4
I_{2} = 1 / 4 (I_{1}) = 1 / 4 (0.00014 W/m^{2})
I_{2} = 0.000035 W/m^{2}
The problem asked for the intensity of the sound wave at your new location (point 2.) No further mathematical solution is necessary.
You are told in the problem to treat the sound as a spherical wave. Some books give the intensity equation for a spherical wave directly. If yours does not, remember that in a spherical wave energy is emitted in all directions. So the power of the wave is spread over a sphere, and the area of a sphere is 4πr^{2}.
In this problem, you were told the intensity of the wave at one point, and so you are able to provide a numerical answer for the intensity of the wave at another point. In many problems, you will not be given this information but you can still compare the points. In this case, you could say that the intensity at your new location is 1/4 the intensity when you were closer to the speaker.
You are not told either your original distance from the speaker (r_{1}) or your new distance from the speaker (r_{2}). However, you are told that you move twice as far away. In other words, you are told that r_{2} = 2r_{1}. This will allow you to divide the unknown quantity, r_{1}, out of the equation.
Power in this equation is the power (energy emitted/time) of the source of the wave. Intensity decreases because this power is spread over a larger and larger area, but the total power of the wave front remains the same. Therefore, I used the same symbol for P at each point, and the unknown value of power could be divided out.
Don’t forget to square the 2! If you double r, r_{2} goes up by a factor of 4.
To divide by a fraction, invert and multiply:
I_{2} / I_{1} = [P / (4πr^{2})]_{2} / [P/(4πr^{2})]_{1}
I_{2} / I_{1} = P / (4πr_{2}^{2}) x (4πr_{1}^{2})/P
I_{2} / I_{1} = P(4πr_{1}^{2}) / P(4πr_{2}^{2})
Another way to solve this kind of problem is to use the information at the “given” point (in this case, your initial location) to solve for the unknown quantities, and to plug that information into the point of interest. For this problem, you could not solve for P and r_{1} directly, but you could solve for the quantity P/r_{1}^{2}. It is perfectly fine to solve the problem that way but does take more time. It is also good to build your ratio reasoning skills, as they often allow you to judge the accuracy of information without fully working a problem.
A spherical wave is a wave that is emitted in all directions. As the wave moves away from its source, its energy spreads spherically around the source and the size of the sphere grows larger as the wave propagates. Because intensity is defined as Power/Area, and because the surface area of a sphere is 4πr^{2}, the intensity of a spherical wave is related to the distance from the source by I = P/4πr^{2}.
Ratios are a very common way to solve definition problems for which you don’t have complete information. As long as you know information about one point, you can solve for information about another point just by taking the ratio of the two definitions. This will be shown in detail in the solution to this problem. This approach can be used for many definitions and not just for intensity.
It is true that I in the equation above is the intensity of a sound wave. However, this equation does not relate intensity to the distance from the source of the sound. Instead, it relates the wave property of intensity (I) to your perception of loudness (I(db)).
See Q & A's Below
A spherical wave is a wave that is emitted in all directions. As the wave moves away from its source, its energy spreads spherically around the source and the size of the sphere grows larger as the wave propagates. Because intensity is defined as Power/Area, and because the surface area of a sphere is 4πr^{2}, the intensity of a spherical wave is related to the distance from the source by I = P/4πr^{2}.
Ratios are a very common way to solve definition problems for which you don’t have complete information. As long as you know information about one point, you can solve for information about another point just by taking the ratio of the two definitions. This will be shown in detail in the solution to this problem. This approach can be used for many definitions and not just for intensity.
It is true that I in the equation above is the intensity of a sound wave. However, this equation does not relate intensity to the distance from the source of the sound. Instead, it relates the wave property of intensity (I) to your perception of loudness (I(db)).
You are told in the problem to treat the sound as a spherical wave. Some books give the intensity equation for a spherical wave directly. If yours does not, remember that in a spherical wave energy is emitted in all directions. So the power of the wave is spread over a sphere, and the area of a sphere is 4πr^{2}.
To divide by a fraction, invert and multiply:
I_{2} / I_{1} = [P/(4πr^{2})]_{2} / [P/(4πr^{2})]_{1}
I_{2} / I_{1} = P / (4πr_{2}^{2}) x (4πr_{1}^{2})/P
I_{2} / I_{1} = P(4πr_{1}^{2}) / P(4πr_{2}^{2})
Another way to solve this kind of problem is to use the information at the “given” point (in this case, your initial location) to solve for the unknown quantities, and to plug that information into the point of interest. For this problem, you could not solve for P and r_{1} directly, but you could solve for the quantity P/r_{1}^{2}. It is perfectly fine to solve the problem that way but does take more time. It is also good to build your ratio reasoning skills, as they often allow you to judge the accuracy of information without fully working a problem.
I_{2} / I_{1} = [P/(4πr^{2})]_{2} / [P/(4πr^{2})]_{1}
I_{2} / I_{1} = r_{1}^{2} / (2 r_{1})^{2}
I_{2} = 1/4 (I_{1}) = 0.000035 W/m^{2}
There are two parts to understanding this problem. The first is just the relationship between intensity and distance. For a wave that spreads out spherically, intensity decreases with distance squared and so you expect to find intensity down by a factor of four when you double the distance.
The second thing to understand in this problem is the usefulness of solving problems with ratios. This type of problem frequently occurs on standardized exams such as the GRE and MCAT. Whenever you are given limited information about two points, and especially if you are asked to compare the two points, it is likely that you can solve the problem most efficiently by taking a ratio of the appropriate equation (usually a definition) at the two points.