Question

Sound intensity varies inversely as the square of the distance from the sound source. If you are in a movie theater and you change your seat to one that is nine times as far from the? speakers, how does the new sound intensity compare with that of your original? seat

Answer 1

$\bf \qquad \qquad \textit{inverse proportional variation} \\\\ \textit{\underline{y} varies inversely with \underline{x}}\qquad \qquad y=\cfrac{k}{x}\impliedby \begin{array}{llll} k=constant\ of\\ \qquad variation \end{array}\\\\ -------------------------------\\\\ \stackrel{\textit{\underline{s} ound intensity varies inversely as the square of the \underline{d}istance from source}}{s=\cfrac{k}{d^2}}$

so, let's say, you have an original distance of say "x" meters, namey d = x, and then you move away and change it to 9 times as much, namely d = 9x, let's check.

$\bf \stackrel{\textit{original distance}}{s=\cfrac{k}{(x)^2}}\qquad \qquad \stackrel{\textit{new distance}}{s=\cfrac{k}{(9x)^2}}\implies s=\cfrac{k}{(9^2x^2)}\implies s=\cfrac{k}{81x^2} \\\\\\ s=\cfrac{k}{x^2}\cdot \cfrac{1}{81}\impliedby \textit{notice, the new sound intensity is }\frac{1}{81}\textit{ of the original}$

Answer 2

Hello.

The further away from the speakers you are the less intense the sound will be.

 If sound intensity varies inversely with the square of the distance, you can represent this as: 
I = 1/d^2 

If you double the distance: 
I = 1/(2d)^2 
I = 1/4d^2 
I = 1/4d^2 

so by doubling the distance, you have 1/4th the intensity.

Have a nice day