Calculating Sound Pressure Level As Distance Changes

[lordterrin]

Homework Statement

The sound pressure level is 80db at a distance of 30 meters. What is the dBSPL at 60 meters?

Homework Equations

My book has provided me the following formula:

dBSPL = dBr-20log(di/dr) - but it has not explained what "di" and "dr" are.

The Attempt at a Solution

I've been looking into this for about two hours, and I'm very confused by what I've read both in my textbook, and online. I understand the concept that Intensity = Energy / (Time * Area), and the concept that as sound travels further away from its source, there is an inverse square relationship between distance and intensity - but I do not understand how to write this out in a formula, or truly how this "inverse square relationship" works. My textbook has the following graph showing distance vs. intensity:

1m = 160 Intensity
2m = 40 Intensity
3m = 17.8 Intensity
4m = 10 Intensity

but I can't figure out how they got there. Mainly, because if the formula is I = E / ( T * A), all I'm being shown is I and A - how do I solve this without knowing E or T?

I'm very frustrated after spending about 2 hours on this. My textbook is useless at actually explaining this to me. I don't just want an answer - I want to understand this!

From what I can gather, I may be correct, or totally off-base with the actual ANSWER to the problem. Either way, I still don't actually understand it.

Using the formula above:

dBSPL = dBr - 20log(di/dr), I think that it goes like this
= 80 - (20 * log(di/dr))
= 80 - (20 * log(60/30))
= 80 - (20 * log(2))
= 80 - log2(20)
= 80 - 4.3219
= 75.67

So - if the sound pressure level at 30m is 80dB, at 60m it would be 75.67dB.

 

[voko]

##d_i## and ##d_r## are two distances from the source. ##dB_r## is the sound pressure level (SPL) at distance ##d_r##. I would expect that the quantity on the left hand side, denoted as dBSPL, should have been denoted as ##dB_i## for consistency. The equation relates SPL at two distances.

Regrading the other formula, the key concept that you may be missing is that all this decibel business deals with ratios. Say we have intensity ##I_0## and we call it our "base" intensity. Then measure some other intensity ##I_1## and then we compare the two by forming the ratio $$I_1 \over I_0$$ Then we take the logarithm of the ratio, and then we scale it with some fancy 20 factor.

Another key thing is that if we measure the intensity of the same wave but at different distances, we know that it carries the same amount of energy in the same period of time. Can you see why? And what that means about the intensities?

 

[lordterrin]

Another key thing is that if we measure the intensity of the same wave but at different distances, we know that it carries the same amount of energy in the same period of time. Can you see why? And what that means about the intensities?

I thought quite a lot about this prior to posting actually, and I tried to write this down in a formula, but I got stuck - because my book shows Intensity as:

Intensity = Energy / (Time * Area)

In the question from my textbook, no Energy or Time is given, so how do I figure out a change when all I have is Intensity and Area?

It makes perfect sense to me that as the distance from the initial source increases, it carries an initial amount of energy over an increasing distance, thus lessening what I might say is "the perceivable power of that energy" (a sound is weaker as you get further away from it,) but your quote above has lost me a bit in regards to the "same period of time" bit. It seems that from an initial source, a sound takes longer to get 60m than it would to get 30m, but the amount of energy remains the same throughout - it's just spread out over a greater distance so it's weaker at 60m than it is at 30m.

So to answer your question, I guess no, I don't see why a wave carries the same amount of energy in the same period of time. I also don't see what that means about intensities.

(It's 12:30am here, I'm going to bed - but will look first thing in the morning - in case you wonder about the gap in response time.)

 

[voko]

It makes perfect sense to me that as the distance from the initial source increases, it carries an initial amount of energy over an increasing distance, thus lessening what I might say is "the perceivable power of that energy" (a sound is weaker as you get further away from it,) but your quote above has lost me a bit in regards to the "same period of time" bit. It seems that from an initial source, a sound takes longer to get 60m than it would to get 30m, but the amount of energy remains the same throughout - it's just spread out over a greater distance so it's weaker at 60m than it is at 30m.

No, this is not correct. Energy in a wave is always localised in time and space, it is not distributed over all the places where the wave has been in the past, it is only where it is now. To make this clear, picture a "parcel" of energy in a wave as a spherical shell of some finite thickness. The thickness is the distance between the "front" and the "rear" of the parcel, and the distance corresponds to the duration of the parcel in time (wave speed times duration). This parcel propagates radially outward. The amount of energy in it is the same (conservation of energy), but, because the radii of the parcel grow, the volume of the parcel grows, so the "density" of energy in the parcel decreases. The "density" of energy is intensity.

 

[lordterrin]

##d_i## and ##d_r## are two distances from the source. ##dB_r## is the sound pressure level (SPL) at distance ##d_r##. I would expect that the quantity on the left hand side, denoted as dBSPL, should have been denoted as ##dB_i## for consistency. The equation relates SPL at two distances.

Regrading the other formula, the key concept that you may be missing is that all this decibel business deals with ratios. Say we have intensity ##I_0## and we call it our "base" intensity. Then measure some other intensity ##I_1## and then we compare the two by forming the ratio $$I_1 \over I_0$$ Then we take the logarithm of the ratio, and then we scale it with some fancy 20 factor.

Okay - what you said (in your last post) makes sense - I was just explaining it in a way that made sense to me, (although it was wrong.)

Judging by what you said here about ##d_r## - am I right in how I built this equation?

dBSPL = dBr - 20log(di/dr), I think that it goes like this
= 80 - (20 * log(di/dr))
= 80 - (20 * log(60/30))
= 80 - (20 * log(2))
= 80 - log2(20)
= 80 - 4.3219
= 75.67

Thinking about what you said about intensities, I guess the dBr is really meaningless until it is compared to a distance, right? 80 decibels can be incredibly loud if I'm 1 inch away from the source, or incredibly weak if it's in outer space and I'm on earth.

Can you explain why we take 20 * log*(di/dr)? I don't get why we multiply the log of (di/dr) by 20.

 

[voko]

The logarithm in the decibel formula is ##\log_{10}##, not ##\log_2##, and ##20 \log_{10} 2 \ne \log_2 20##. Correct that and you should be fine.

I am not completely sure what you mean by 80 dB being meaningless. It certainly means that at the location where it was measured, the ratio with the base level was 80 dB. To deduce the ratio at some other location, of course, you would have to know the distances from both locations to the source.

As for the 20 factor, it is really 2 times 10. The 2 in it has to do with the inverse square law we discussed earlier (we take the ratio of distances squared, and the square becomes factor 2 in front of the logarithm). The factor 10 is arbitrary, but it is because of it that the unit is called the deci-bel. There is also a (much) less frequently used unit bel, which does not have the 10 factor.

 

[lordterrin]

The logarithm in the decibel formula is ##\log_{10}##, not ##\log_2##, and ##20 \log_{10} 2 \ne \log_2 20##. Correct that and you should be fine.

Why is it ##\log_{10}## ? I don't see that anywhere in the formula the professor gave us (up top)?

 

[voko]

Well, it is :) Either trust me, or look here: http://en.wikipedia.org/wiki/Decibel or anywhere else.