"Sound Intensity Decreases with Distance

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Homework Help Overview

The discussion revolves around the intensity of sound and its relationship with distance from the source, specifically examining how sound intensity measured in decibels (dB) changes as one moves from 1 meter to 10 meters away from the sound source.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the relationship between sound intensity and distance, questioning how to relate decibels to power ratios. There is discussion about the equations for intensity and the surface area of spheres, as well as the logarithmic nature of decibel calculations.

Discussion Status

Some participants have provided guidance on using the decibel formula for power ratios and have clarified the concept of power flux in relation to sound. There is an ongoing exploration of the mathematical relationships involved, with no explicit consensus reached yet.

Contextual Notes

Participants are working within the constraints of a homework problem that requires understanding the principles of sound intensity and decibel calculations, with some assumptions about the reference power flux and the geometry of sound propagation being discussed.

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Homework Statement


At 1 m away from the source of a sound, the intensity of the sound is 90 dB. At 10 m away the intensity is…?

The answer is 70 dB but I don't know how...

Homework Equations



I = Power/Area


The Attempt at a Solution


My thought was to set up the equations:

90dB = P/4pi
XdB = 4pir^2

I'm not sure where to go from here.
 
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Aoiumi said:

Homework Statement


At 1 m away from the source of a sound, the intensity of the sound is 90 dB. At 10 m away the intensity is…?

The answer is 70 dB but I don't know how...

Homework Equations



I = Power/Area


The Attempt at a Solution


My thought was to set up the equations:

90dB = P/4pi
XdB = 4pir^2

I'm not sure where to go from here.

You are on the right track.

What is the ratio of the surface area of the two spheres?

And what is the equation for power in dB based on the ratio of intensities?
 
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The ratio of the surface area of the two spheres would be

4pi/4pi(10)^2 or
1/100

I'm not sure how to set up the quation for power in dB based on the ratio of intensities

I0/I = (p1/a1)/(p2/a2)?
 
Aoiumi said:
The ratio of the surface area of the two spheres would be

4pi/4pi(10)^2 or
1/100

I'm not sure how to set up the quation for power in dB based on the ratio of intensities

I0/I = (p1/a1)/(p2/a2)?

You use dB to handle large ratios.

For voltage, V(dB) = 20 * log(V/Vo) where Vo is some reference voltage

For power P(dB) = 10 * log(P/Po) where Po is some reference power

So for this question, you will be using the 2nd equation...
 
First, allow me to offer some general guidance regarding decibels.

Decibels always imply power ratios. Always. No matter what.

The decibel is defined, relating the ratio of two powers P2 and P1 as:

10 \log_{10} \left( \frac{P_2}{P_1} \right)
So one might ask, "well then how can we use decibels with voltage? Voltage isn't power," or "where does the '20' come from when we use ratios of voltages?"

The answer is that we make the pseudo-assumption that the input and output impedances of a given circuit are equal; i.e., Rin = Rout = R. (By the way, this is a realistic pseudo-assumption, since it is often the case for circuits that are optimized for maximum power transfer, signal to noise ratio, etc.)

Also we know that the power through a resistor is V2/R.

Plugging that into our decibel equation, and noting that log(x2) = 2logx,

10 \log_{10} \left( \frac{ \frac{V_2^2}{ R}}{ \frac{V_1^2}{ R}} \right) = 10 \log_{10} \left( \frac{V_2^2}{V_1^2} \right) = 10 \log_{10} \left( \left[ \frac{V_2}{V_1} \right]^2 \right) = 20 \log_{10} \left( \frac{V_2}{V_1} \right)

----------------------------

Now allow me to move on to something more specific: Sound/acoustics.

Sound decibels are defined in terms of power flux* ratios.

*(Power flux is power through a unit area -- you may think of the unit area as the cross sectional area of your ear canal if it helps visualize it -- or if you want to stick with SI units, the sound power propagating through a square meter.)

The reference power flux is typically defined as the power flux that corresponds to the human threshold of hearing.

If we define the reference power flux, that of the threshold of human hearing, as \Phi_{ref}, and the power of the source as P, then we know from the problems statement that

90 \ \mathrm{dB} = 10 \log_{10} \left( \frac{\frac{P}{A_1}}{\Phi_{ref}} \right)
where A_1 is the surface area of a circle with a 1 meter radius. So now the question is, what is

10 \log_{10} \left( \frac{\frac{P}{A_{10}}}{\Phi_{ref}} \right) \ ,
where A_{10} is the surface area of a circle with radius 10 meters?

Hint: There is an easy way to do this such that you won't even need a calculator to solve this problem (you might even be able to solve it in your head). Do this by noting that log(x/y) = logx - logy, and asking yourself, Does the area of a circle increase proportionally with the radius, or does it increase proportionally with the square of the radius?" :wink:
 
Last edited:
Thanks for the clarification, collinsmark. Nice post :smile:
 
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