Sound waves - Destructive Interference

1. Apr 8, 2013

Saitama

1. The problem statement, all variables and given/known data
Two identical loudspeakers are located at points A & B, 2m apart. The loudspeakers are driven by the same amplifier (coherent and are in the same phase). A small detector is moved out from point B along a line perpendicular to the line connecting A & B. Taking speed of sound in air as 332 m/s, find the frequency below which there will be no position along the line BC at which destructive interference occurs.

2. Relevant equations

3. The attempt at a solution
I am not sure how would I approach this problem. I started with calculating the path difference when the detector is at a distance x from its initial position,
$$\Delta x=\sqrt{4+x^2}-x$$
For destructive interference,
$$\sqrt{4+x^2}-x=\left(n+\frac{1}{2} \right)\lambda$$
where $\lambda$ is the wavelength of the wave.

I don't know how should I proceed from here.

Any help is appreciated. Thanks!

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2. Apr 8, 2013

ehild

Think:
What does lowest frequency mean for the wavelength?
How does the pathlength - difference change if the detector moves along x?

ehild

3. Apr 8, 2013

Saitama

Lowest frequency means highest wavelength. For that should I substitute n=0? And then I will get $\lambda$ as a function of x. So should I find maximum of this function?

4. Apr 8, 2013

ehild

Yes, do it...

ehild

5. Apr 8, 2013

Saitama

I tried that but had no luck. Substituting n=0,
$$\lambda=2(\sqrt{4+x^2}-x)$$
Differentiating w..r.t x
$$\frac{d\lambda}{dx}=2(\frac{x}{\sqrt{4+x^2}}-1)=0$$
This equation has no solution for x. :(

EDIT: Looks like the function has its maximum value at x=0. And this gives me the right answer. Thanks ehild!

6. Apr 8, 2013

ehild

There is no local maximum, but the function decreases with x, and x≥0. The function takes the maximum at the boundary.

ehild

7. Apr 8, 2013

Saitama

Yes I realised it when I examined the derivative. :)