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So, there is a sound at $S$, whose intensity $I$ obeys the inverse square law ($I \sim \frac{1}{x^2}$). At point $P$, at a distance $r$ from $S$, the air molecules oscillate with an amplitude of $8μm$. Point $Q$ is at a distance of $2r$ from $S$. What is the amplitude of the air molecules at $Q$? What is the relationship between amplitude and distance?

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The amplitude $A$ is related to the intensity $I$ as follows: $I\sim A^2$ – Raskolnikov Mar 30 '11 at 18:11
So, er, $A^2 \sim \frac{1}{x^2}$ and therefore $A \sim \frac{1}{x}$? – Electro Mar 30 '11 at 18:17
up vote 1 down vote accepted

Yes, Electro, the intensity $I$ - and energy density $T_{00}$ and similar quantities - is proportional to the squared amplitude, $$ I \sim A^2 $$ Because the intensity must go as $$ I \sim \frac{1}{r^2} $$ as the energy gets spread over the sphere of area $4\pi r^2$, it follows that $$ A \sim \frac{1}{r}.$$ See e.g. the $1/r$ factor in this formula:

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