What is the relationship between the magnitude of a sound wave and distance?

In summary, the magnitude of a wave emanating from an image as a function of distance would be described by the following equation: y = \frac {1} {4 \pi r}
  • #1
dimensionless
462
1
In basic acoustic theory a sound source reflects across a boundry in a fasion similar to light. The magnitude of the reflected wave can be determined by adjusting it for the reflection loss and the distance traversed.
I am looking at a scientific paper from the 70's. In it they use the following equation to caculate the magnitude of a wave emanating from an image x meters away.
[tex]
y = \frac {1} {x}
[/tex]
My intuition would tell me that the magnitude of a wave emanating from an image as a function of distance would be described by the following:
[tex]
y = \frac {1} {4 \pi x^2}
[/tex]
Does anyone know if the former equation is really correct? Why is the inverse square law not used?
 
Last edited:
Science news on Phys.org
  • #2
dimensionless said:
In basic acoustic theory a sound source reflects across a boundry in a fasion similar to light. The magnitude of the reflected wave can be determined by adjusting it for the reflection loss and the distance traversed.
I am looking at a scientific paper from the 70's. In it they use the following equation to caculate the magnitude of a wave emanating from an image x meters away.
[tex]
y = \frac {1} {r}
[/tex]
My intuition would tell me that the magnitude of a wave emanating from an image as a function of distance would be described by the following:
[tex]
y = \frac {1} {4 \pi r^2}
[/tex]
Does anyone know if the former equation is really correct? Why is the inverse square law not used?

the magnitude of any radiation intensity, that is something measured in watts/m^2 must be inverse square (assuming no transmission loss) to satisfy the conservation of energy axiom.

however for sound, the intensity of radiation is the product of the RMS pressure, which is 1/r for a spherically expanding wavefront, and the RMS of the in-phase component particle velocity, which is also 1/r. it turns out that the quadrature component of the particle velocity (that is the component that is 90o out of phase with the pressure wave) is 1/r^2, but that component does not contribute to power transmission. so if both pressure and in-phase particle velocity are 1/r, then their product is 1/r^2 or inverse square.

there is another thread here called Inverse Square Relationship that you might want to check out. if not, do check out

http://en.wikipedia.org/wiki/Inverse_square_law

http://en.wikipedia.org/wiki/Flux

http://en.wikipedia.org/wiki/Gauss's_law

the first link talks a little about acoustic waves and 1/r vs. 1/r^2
 
  • #3
So is the pressure of the sound wave [tex]P = \frac {1} {4 \pi r}[/tex]?? Or is it [tex]P = \frac {1}{r}[/tex]?? The references don't seem to make it terribly clear.
 

Related to What is the relationship between the magnitude of a sound wave and distance?

1. What is the magnitude of acoustical image?

The magnitude of acoustical image refers to the intensity or strength of the sound waves that make up the image. It is a measure of the amplitude of the sound waves and is often represented on a scale, such as decibels (dB).

2. How is the magnitude of acoustical image measured?

The magnitude of acoustical image is typically measured using specialized equipment such as a microphone, hydrophone, or acoustic camera. These devices capture the sound waves and convert them into electrical signals, which can then be analyzed to determine the magnitude of the image.

3. What factors can affect the magnitude of acoustical image?

The magnitude of acoustical image can be influenced by several factors, including the distance between the sound source and the image, the type and characteristics of the sound source, and the properties of the medium through which the sound travels. Other factors such as reflections, interference, and absorption can also impact the magnitude of the image.

4. How does the magnitude of acoustical image relate to image clarity?

The magnitude of acoustical image is directly related to image clarity. A higher magnitude generally indicates a stronger or clearer image, while a lower magnitude can result in a weaker or less defined image. However, other factors such as the frequency of the sound and the resolution of the imaging equipment also play a role in image clarity.

5. Can the magnitude of acoustical image be changed?

The magnitude of acoustical image can be altered by adjusting the properties of the sound source, such as its volume or frequency. It can also be affected by changing the environment or using sound-absorbing materials to reduce reflections. However, the magnitude of the image ultimately depends on the properties of the sound source and the medium through which it travels.

Similar threads

Replies
31
Views
856
  • Mechanics
Replies
5
Views
1K
  • General Math
Replies
2
Views
2K
Replies
2
Views
1K
  • Calculus and Beyond Homework Help
Replies
1
Views
2K
Replies
3
Views
2K
  • Introductory Physics Homework Help
Replies
29
Views
1K
  • Introductory Physics Homework Help
Replies
6
Views
4K
  • Introductory Physics Homework Help
Replies
1
Views
1K
  • Mechanics
Replies
20
Views
3K
Back
Top