# Calculating Sound Pressure Level As Distance Changes

**1. The problem statement, all variables and given/known data**

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

**2. Relevant equations**

My book has provided me the following formula:

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

**3. 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.

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