Power and Intensity for a Sound Wave

In summary: I understand it much better now.In summary, the conversation discussed the equations for power and intensity of a sound wave, and the relationship between them and the cross-sectional area of the wave. It was clarified that the area in the equations is where the power or intensity is being measured and can be at the source or other locations. It was also noted that for spherical waves, the intensity and pressure/displacement graphs would follow an inverse-square law with distance from the source. The conversation ended with a better understanding of the concept.
  • #1
Gear300
1,213
9
The power for a sound wave is give by P = 1/2*p*A*(w*s)^2*v...in which p is density (rho), A is cross-sectional area, w is angular frequency, s is maximum displacement (amplitude), and v is speed of propagation. The intensity is given by P/A...the intensity for a spherical sound wave is supposed to decrease over time (according to experience)...but I'm not seeing it in the equations. The power is proportional to the area over some region and the intensity is inversely proportional to the area over some region. Therefore...the Intensity at a given point shouldn't depend on A (the area)...that would make the intensity constant throughout the wave. Where am I wrong in the reasoning?

Oh wait...never mind...the source determines the power...I shift to a new question: the A (area) in the equations is the cross-sectional area immediately "touching" the source, right?
 
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  • #2
No, A is the area at the place where power or intensity is being measured.
 
  • #3
So in that case...if a drum is the source of a sound...then the area A would be the surface area of the oscillating surface of the drum, correct?
 
  • #4
ಠ_ಠ ?
 
  • #5
A is the area at wherever the sound power is being calculated or measured. It could be at the drum's surface, it could also be somewhere else.
 
  • #6
Wait...how could it also be somewhere else? Isn't it where the surface of the source intercepts the medium? Is it an arbitrary value?
 
  • #7
ಠ_x?...the area A I'm referring to is the one in the equation for sound wave power: P = 1/2*p*A*(w*s)^2*v
 
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  • #8
Do sound waves have power only right at the source? Or do they also have power wherever they travel?
 
  • #9
Redbelly98 said:
Do sound waves have power only right at the source? Or do they also have power wherever they travel?

They have power wherever they travel...but how does that fit in the equation if density, angular velocity, maximum displacement, and velocity are constant (it would only leave area, A, as variable)?
 
  • #10
Maximum displacement will be different, decreasing as you move farther from the source.

edit:
Another suggestion: is A defined in the book or article you got the equation from? Perhaps they are only concerned with power at the source, and that's all that you have to be concerned with ... even though we could also calculate/measure power at locations away from the source if we wish to.
 
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  • #11
Redbelly98 said:
Maximum displacement will be different, decreasing as you move farther from the source.

edit:
Another suggestion: is A defined in the book or article you got the equation from? Perhaps they are only concerned with power at the source, and that's all that you have to be concerned with ... even though we could also calculate/measure power at locations away from the source if we wish to.

oh...that makes sense...
 
  • #12
Does this mean that for spherical waves, the pressure and displacement graphs are damped?
In the book, A is not explicitly defined. They come up with a proof for the energy and power equations using a model, in which the area is taken at the source...thing is that in this model, the area stays constant (sound propagating through a cylinder). They then defined intensity for this model, which would also stay constant due to the constant area...then they generalized the situation for spherical waves with I = P(avg)/A. Looks like I wasn't understanding the concept well.
 
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  • #13
Gear300 said:
Does this mean that for spherical waves, the pressure and displacement graphs are damped?

Yes. Intensity will follow an inverse-square law with distance from the source.
Pressure and displacement amplitudes are proportional to square root of intensity, and hence are inversely proportional to distance from the source.

In the book, A is not explicitly defined. They come up with a proof for the energy and power equations using a model, in which the area is taken at the source...thing is that in this model, the area stays constant (sound propagating through a cylinder). They then defined intensity for this model, which would also stay constant due to the constant area...then they generalized the situation for spherical waves with I = P(avg)/A.

Okay.
"A" would be the surface area of a sphere, at a distance "r" from the source.
So A = 4 pi r2, giving an inverse-square relation for I=P/A
 
  • #14
Alright...thanks for clarifying things
 

Related to Power and Intensity for a Sound Wave

1. What is the difference between power and intensity for a sound wave?

Power and intensity are both measures of the energy carried by a sound wave, but they differ in how they are measured. Power is the rate at which energy is transferred by the sound wave, while intensity is the amount of energy passing through a unit area perpendicular to the direction of the wave. In other words, power is a measure of the amount of energy produced by the sound source, while intensity is a measure of the strength of the sound wave at a specific point in space.

2. How is power related to the amplitude of a sound wave?

The power of a sound wave is directly proportional to the square of its amplitude. This means that doubling the amplitude will result in four times the power, while tripling the amplitude will result in nine times the power. This relationship is known as the inverse square law, and it applies to both power and intensity of a sound wave.

3. What units are used to measure power and intensity of a sound wave?

The SI unit for power is watts (W), while the SI unit for intensity is watts per square meter (W/m²). However, in acoustics, it is common to use decibels (dB) to express both power and intensity, which is a logarithmic scale that allows for a wider range of values to be easily represented.

4. How does the distance from a sound source affect power and intensity?

As a sound wave travels through space, its power remains constant, but its intensity decreases as it spreads out over a larger area. This means that the farther you are from a sound source, the lower the intensity will be. The relationship between distance and intensity follows the inverse square law, so doubling the distance will result in one-fourth the intensity, while tripling the distance will result in one-ninth the intensity.

5. Can you measure the power and intensity of a sound wave without specialized equipment?

Technically, yes, you can measure the power and intensity of a sound wave using basic equipment such as a microphone and a sound level meter. However, for accurate and precise measurements, specialized equipment such as an anechoic chamber and calibrated microphones are necessary. This is because there are many factors that can affect the measurement of power and intensity, such as environmental noise and microphone sensitivity.

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