# Interference between two speakers driven by the same amplifier

Two speakers are driven by the same amplifier with a frequency of 1380Hz. The two speakers are 4m apart. An observer, originally at the position of one of the speakers, starts moving away along a line perpendicular to the line connecting the two speakers. Calculate the number of sound minima the observer will hear while the observer is moving to a spot 3m from the speakers. Assume the speed of sound is 343m/s.

The solution first calculates the path difference when the observer is at one of the speakers.

$\text{path difference}=\Delta r=\sqrt{4^2+0^2}-0^2=4$

And equate $\Delta r=m \lambda$ where $m$ is an integer and $\lambda$ is the wavelength. $\lambda=\frac{343m/s}{1380Hz}=0.25m$, then $m=16$.

Then the solution do the same thing, equate the path difference and $n\lambda$ when the observer is at 3m away. $\Delta r=\sqrt{4^2+3^2}-3=2=n\lambda \Rightarrow n=8$

Then the solutions says $\Delta r=(n+\frac{1}{2})\lambda \Rightarrow 8<n+\frac{1}{2}<16 \Rightarrow \text{eight minima}$

I do not understand what the solution is doing.

• Why does it use the formula for constructive interference $\Delta r=m\lambda$. Isn't this a problem of destructive interference?
• For the inequality $8<n+\frac{1}{2}<16$, why is the inequality strict? I can see that the path difference must be between 2 and 4. That is, $2 \leq(n+\frac{1}{2})\lambda \leq 4 \Rightarrow 8 \leq n+\frac{1}{2} \leq 16$ This will give 10 minima. But why is it strict? Is there something got to do with the constructive interference formula?

I'm sorry if my description is not clear. It is because I don't really understand the physics behind this problem since I first encountered it.

• What do you think the solution should be? – sammy gerbil Feb 3 '18 at 9:03

At the speaker the path difference is $16\lambda$ and when $3 \rm m$ away the path difference is $8 \lambda$.
So the lowest order path difference for a minimum is $8.5 \lambda$ and the highest order for a minimum is $15.5 \lambda$.
If you count then that means that there are $8$ minima in total and they are $15.5,\,14.5,\,13.5,\,12,5,\,11.5,\,10.5,\,9.5$ and $8.5$.
All these minima satisfy $$15.5\lambda \le (n + \frac 12)\lambda \le 8.5 \lambda$$ where $n$ is an integer with $n = 15,\, 14,\,13,\,12,\,11,\,10,\,9,\,8$.
This inequality has been put another way $$16\lambda < (n + \frac 12)\lambda < 8 \lambda$$ knowing that there are no minima between $16 \lambda$ and $15.5 \lambda$, and $8.5 \lambda$ and $8 \lambda$