Intensity and Power of Sound

In summary: So I think thatr_2 = r_1 + d makes more sense. I think that the problem wants you to find the distance from the source to the closer listener, given that the closer listener hears a sound with intensity level of 55.8 dB and the more distant listener hears a sound with intensity level of 49.8 dB. If that's the case, then you can use the first two equations and solve for I/I0. Then, you can use the third equation to solve for r1. If you do that, you should find that r1 should be equal to r_1 = \frac{1}{10^{(6/20)}} * d where again d =
  • #1
getty102
38
0

Homework Statement



The sound intensity level of a certain sound source is measured by two listeners located at different positions along a line from the source. The listeners are located on the same side of the source and are separated by 34.8 m. The listener that is closest to the source hears the sound with a sound intensity level of 55.8 dB. The sound intensity level of the sound heard by the more distant listener is 49.8 dB.

Homework Equations



β = 10dBlog10(I/I0)
Iav=Pav/4∏r2

The Attempt at a Solution



β2 - β1 = 20dBlog10(r2/r1)

10^((β21)/20) = r2/r1.

r1/r2 = 1/10^((β21)/20)

r1((1/34.8)+1) = 1/10^((β21)/20)

r1 = 1/10^((β21)/20)/((1/34.8)+1)

This doesn't seem to be working, I'm thinking my math is getting messed up somewhere's along the line.
 
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  • #2
getty102 said:

Homework Statement



The sound intensity level of a certain sound source is measured by two listeners located at different positions along a line from the source. The listeners are located on the same side of the source and are separated by 34.8 m.
I interpret the last statement as saying that the listeners are separated from each other by 34.8 meters. It's kind of ambiguously worded, so I'm not 100% certain. But for the rest of this post I'm going to assume that they are separated from each other by 34.8 meters, such that

r2 = r1 + d

where d = 34.8 meters. r1 is the distance from the source to the closer listener, and r2 is the distance from the source to the more distant listener.

If I'm interpreting this incorrectly, let me know.
The listener that is closest to the source hears the sound with a sound intensity level of 55.8 dB. The sound intensity level of the sound heard by the more distant listener is 49.8 dB.
I'm a little confused now. The problem statement gives a lot of statements, yet no instructions on what it is we're supposed to find. The 'problem' itself hasn't been specified. In other words, what is the question?

Homework Equations



β = 10dBlog10(I/I0)
Iav=Pav/4∏r2

The Attempt at a Solution



β2 - β1 = 20dBlog10(r2/r1)

10^((β21)/20) = r2/r1.
So far so good. :approve:
r1/r2 = 1/10^((β21)/20)
I'm not sure why would want to invert both sides of the equation at this point in the process. I don't see a purpose in that. But, okay...
r1((1/34.8)+1) = 1/10^((β21)/20)
Now you've lost me. :uhh:

According to your above equation compared with the previous one, you substituted something in for r2,

[tex] r_2 = \frac{1}{\frac{1}{d} + 1} [/tex]

I don't think that's justified. From the way I interpret the problem statement, r2 = r1 + d.
 

What is the difference between intensity and power of sound?

Intensity and power of sound are two related but distinct concepts in acoustics. Intensity refers to the amount of energy that passes through a unit area per unit time, while power refers to the rate at which energy is transferred. In simpler terms, intensity measures the strength or loudness of a sound, while power measures the amount of energy that is produced by the sound source.

How is intensity of sound measured?

Intensity of sound is typically measured in decibels (dB), a logarithmic unit that compares the sound level to a reference level. The most commonly used reference level is 0 dB, which is considered to be the threshold of human hearing. As the intensity of sound increases, the decibel level also increases, with 10 dB representing a ten-fold increase in intensity.

What factors affect the intensity of sound?

The intensity of sound is affected by several factors, including the distance from the sound source, the size and shape of the sound source, and the medium through which the sound is traveling. As sound waves travel through a medium, they spread out and decrease in intensity, meaning that the farther away you are from the sound source, the lower the intensity of the sound will be.

How does the intensity of sound impact human hearing?

The intensity of sound can have a significant impact on human hearing. Prolonged exposure to sounds above 85 dB can cause permanent damage to the delicate structures of the inner ear, leading to noise-induced hearing loss. It is important to protect your ears from high-intensity sounds, especially in noisy environments such as concerts or construction sites.

What are some practical applications of understanding intensity and power of sound?

Understanding intensity and power of sound has many practical applications in various fields, including music production, engineering, and medicine. In music production, understanding the intensity of sound is crucial for creating balanced and pleasing mixes. In engineering, it is essential for designing and testing soundproofing materials. In medicine, it is used to diagnose and treat hearing loss and other auditory disorders.

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