The loudness of a sound is inversely proportional to the square of your distance from the source of the sound. if your friend is right next to the speakers at a loud concert and you are four times as far away from the speakers, how does the loudness of the music at your position compare to the loudness at your friend's position?

Let loudness be L, distance be d, and k be the constant of variation such that the equation that would best represent the given above is,
L = k/(d^2)
For Case 1,
L1 = k/(d1^2)
For Case 2,
L2 = k/((d1/4)^2)
For k to be equal, L1 = 16L2.
Therefore, the loudness at your friend's position is 16 times that of yours.

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aristocles aristocles · Answer 2 · 2019-08-17T05:04:02+02:00

Answer:

Loudness at our position is 1/16 times smaller than the loudness at your friend position

Explanation:

As we know that the loudness decreases as we move away from the speaker

and the loudness is inversely depending on the square of the distance from the speaker

So here we know

[tex]L = \frac{k}{r^2}[/tex]

so here we can compare two loudness at two different positions

[tex]\frac{L_1}{L_2} = (\frac{r_2}{r_1})^2[/tex]

here we know that

[tex]r_2 = 4 r_1[/tex]

now we know that

[tex]\frac{L_1}{L_2} = 4^2[/tex]

[tex]L_2 = \frac{L_1}{16}[/tex]