Sound waves - Destructive Interference

In summary, by calculating the path difference and setting it equal to n+1/2 times the wavelength, the maximum wavelength can be found by substituting n=0 and finding the maximum of the resulting function. Since the function decreases with increasing x and x≥0, the maximum value is taken at the boundary, which gives the right answer.
  • #1
Saitama
4,243
93

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!
 

Attachments

  • sw1.png
    sw1.png
    1.2 KB · Views: 495
Physics news on Phys.org
  • #2
Think:
What does lowest frequency mean for the wavelength?
How does the pathlength - difference change if the detector moves along x?

ehild
 
  • #3
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?
 
  • #4
Yes, do it...

ehild
 
  • #5
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:
 
  • #6
There is no local maximum, but the function decreases with x, and x≥0. The function takes the maximum at the boundary.

ehild
 
  • #7
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. :)
 

Related to Sound waves - Destructive Interference

1. What is destructive interference in sound waves?

Destructive interference in sound waves occurs when two waves with opposite amplitudes meet and their crests and troughs cancel each other out, resulting in a decrease in the overall amplitude of the wave.

2. How does destructive interference affect the volume of a sound?

Destructive interference can significantly decrease the volume of a sound. When two waves cancel each other out, the resulting wave has a lower amplitude, which translates to a lower volume.

3. What causes destructive interference to occur in sound waves?

Destructive interference in sound waves is caused by the superposition of two waves with opposite amplitudes. This can occur when two sound sources are out of phase with each other or when a sound wave reflects off a surface.

4. Can destructive interference be beneficial in certain situations?

Yes, destructive interference can be beneficial in some situations. For example, it can be used to reduce unwanted noise in a sound system by canceling out specific frequencies.

5. How does the distance between two sound sources affect destructive interference?

The distance between two sound sources can greatly affect destructive interference. The closer the sources are to each other, the more pronounced the interference will be. As the distance between the sources increases, the interference decreases.

Similar threads

  • Introductory Physics Homework Help
Replies
3
Views
728
  • Introductory Physics Homework Help
Replies
7
Views
1K
  • Introductory Physics Homework Help
Replies
7
Views
4K
Replies
3
Views
2K
  • Introductory Physics Homework Help
Replies
2
Views
3K
Replies
3
Views
1K
  • Introductory Physics Homework Help
Replies
19
Views
3K
  • Introductory Physics Homework Help
Replies
20
Views
2K
  • Introductory Physics Homework Help
Replies
5
Views
1K
  • Introductory Physics Homework Help
Replies
15
Views
2K
Back
Top