# Can the intensity of a sound wave in a minimum be greater than intensity in a maximum?

Suppose to have two speakers $A$ and $B$ one in front of the other, producing coherent sound waves with the same frequency $f$ and power $P$. In different positions on the line connecting the speakers there is constructive and distructive interference.

Where the interference is constructive, at distances $r_{A,max}$ from $A$ and $r_{B,max}$ from $B$, the intensity of the resulting wave is $$I_{MAX}=(\sqrt{I_A}+\sqrt{I_B})^2=\frac{P}{4 \pi} \bigg(\frac{1}{r_{A,max}}+\frac{1}{r_{B,max}}\bigg)^2$$

Where the interference is distructive, at distances $r_{A,min}$ from $A$ and $r_{B,min}$ from $B$, the intensity of the resulting wave is $$I_{MIN}=(\sqrt{I_A}-\sqrt{I_B})^2=\frac{P}{4 \pi}\bigg(\frac{1}{r_{A,min}}-\frac{1}{r_{B,min}}\bigg)^2$$

So I can consider the ratio

$$q=\frac{I_{MAX}}{I_{MIN}}=\frac{\bigg(\frac{1}{r_{A,max}}+\frac{1}{r_{B,max}}\bigg)^2}{\bigg(\frac{1}{r_{A,min}}-\frac{1}{r_{B,min}}\bigg)^2}$$

In a exercise with a similar situation I found $q<1$, which means $I_{MAX}<I_{MIN}$ and this seems strange because the intensity in a minimum would be greater than one in a maximum. Is that possible?

It seems to me that the resulting frequency is greater when the distances between speakers are similar, indipendently from the fact that I have a constructive or distructive interference but how can I obtain a lower intensity in a maximum than in a minimum?

I report the exercise to show the problem

An observer $O$ is between two speakers $A$ and $B$ emitting at same frequency and power. The distance between the speakers is $d=2m$ and $O$ is at distance $r_{A,max}=0.5 m$ from $A$ when he hears the maximum. The observer moves of a distance $0.2 m$ and hears a minimum. Find the ratio $q$.

It is not specified if the observer moves towards $A$ or $B$ but, if I consider that it moves towards the nearest speaker, $A$, then I get $q<1$, infact

$$q=\frac{I_{MAX}}{I_{MIN}}=\frac{\bigg(\frac{1}{0.5}+\frac{1}{1.5}\bigg)^2}{\bigg(\frac{1}{0.3}-\frac{1}{1.7}\bigg)^2}=0.94$$