sound intensity follows the inverse square law - therefore work out this question...........?

Question

if the sound intensity at 2 metres from a source is 90dB, what is the sound intensity at 6 metres from the source??

i have worked it out but need to know if it is correct. would you need to use a formula for this calculation?

Answer 1

The distance is 3 times the original. Therefore the sound level is 1/9 the original according to the inverse square law.

Difference in decibels is 10log(base10)(1/9) = -9.542425

Therefore as decibels are logarithmic.

sound intesity = 90dB - 9.542425dB
= 80.457575dB

Answer 2

You don't need a specific formula if you know that the inverse square law applies

Intensity L = x / d^2
this is the inv sq law where L is the sound intensity and d is the distance from the source. x is the constant of proportionality

all you need to do is consider the ratio of the squares of the distances
2^2 / 6^2 = 4/36 = 1/9
therefore sound intensity at 6m = 1/9 * 90 = 10dB

Answer 3

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Answer 4

I2/I1 = (d1/d2)^2
90dB = 10Log(I/Ir)
I/Ir = 10^9 @ 2 m
I/Ir = 4/36*10^9 @ 6 m
I/Ir = 4/36*10^9 = 1.111111*10^8
Log10(I2/Ir) = 8.045757
10Log10(I2/Ir) = 80.458 dB