Sound waves - Destructive Interference

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Saitama
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Homework Statement


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.

Homework Equations


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,
[tex]\Delta x=\sqrt{4+x^2}-x[/tex]
For destructive interference,
[tex]\sqrt{4+x^2}-x=\left(n+\frac{1}{2} \right)\lambda[/tex]
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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ehild said:
Think:
What does lowest frequency mean for the wavelength?
How does the pathlength - difference change if the detector moves along x?

ehild

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?
 
ehild said:
Yes, do it...

ehild

I tried that but had no luck. Substituting n=0,
[tex]\lambda=2(\sqrt{4+x^2}-x)[/tex]
Differentiating w..r.t x
[tex]\frac{d\lambda}{dx}=2(\frac{x}{\sqrt{4+x^2}}-1)=0[/tex]
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! :smile:
 
ehild said:
There is no local maximum, but the function decreases with x, and x≥0. The function takes the maximum at the boundary.

ehild

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