"Sound Intensity Decreases with Distance

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The discussion focuses on calculating sound intensity at different distances, specifically how the intensity decreases from 90 dB at 1 meter to 70 dB at 10 meters. Participants emphasize the importance of understanding the relationship between sound intensity and surface area, noting that intensity decreases with the square of the distance from the source. The relevant equations for power in decibels are highlighted, particularly the formula for power ratios in decibels. The conversation also clarifies that decibels represent power ratios and explains the concept of power flux in relation to sound. Overall, the key takeaway is that sound intensity decreases logarithmically with distance due to the geometry of sound propagation.
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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:
 
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Thanks for the clarification, collinsmark. Nice post :smile:
 
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Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
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