Question

Jim halves the distance between himself and a sound source. What is the change in decibels of the sound he hears?

+6 dB

−2 dB

+2 dB

−6 dB

Answer 1

First, you must understand the change in the intensity of the sound after the distance changes. The distance has been halved, so the sound intensity has been changed by 1/4, correct?

Now from the intensity, find the decibel value. Recall the decibel is a measure of intensity using the log scale. Hence, to find the change in decibels, β, you must apply the change in intensity using the log scale.

\(β=10log(\frac{I}{I_o})\)
\(β=10log(\frac{1}{4})\)
Once you find the value of β, or dB, you may need to round it slightly to find the answer here. Let me know if you have any questions.

Answer 2

so it would be -6 dB?

Answer 3

You just solve for B right?

Answer 4

Yes I got -6 dB as well but now I'm re-evaluating the choice but now I realize that the distance has decreased, just a sec

Answer 5

yeah I just noticed by mistake after I came up with fake numbers

If the distance was halved, then it would be 4 times stronger, so you would do 10log(4) and get +6 not -6

Answer 6

The wording confused me here for a moment, but I gave myself fake numbers and used the Inverse Square Law

\(\frac{I_2}{I_1}=(\frac{d_1}{d_2})^2\)

I gave myself fake numbers, I let the first distance d1 = 30 and d2 = 15, then you square and get I2/I1 = 4. β=10log(4) ≈ 6dB

Answer 7

but would it be positive 6 if Jim quadruples the distance between himself and a sound source?

Answer 8

wouldnt it be*

Answer 9

No it would not be quadrupled, it would be halved. Try plugging in random values into the inverse square law. Quadrupling the distance would decrease the dB by -12 dB. Now REDUCING quadruple the distance would increase the dB by 12 dB.

But here it is saying halved. Like I said, try and plug in random values for d1 and d2 in the inverse square law. The original distance is, say, 40 meters, and now the final distance, d2, is 20 meters. This gives you an I/Io ratio of 4, taking 10log(4) = 6. Works with any distance as long as it's a half ratio.

Answer 10

I just tried plugging in values and it works! It would +6, I was getting confused with the terms but once I reread it I understood it. thank you for explaining that for me. That makes sense as well for the quadrupling which would make that -12 dB.