Is the Path Difference Calculation Correct for Destructive Interference?

In summary, the conversation discusses the concept of destructive interference occurring at specific points (0.5(lambda), (3/2)(lambda), etc.), and the attempt at solving for path difference to achieve this interference. The author shares their attempt at finding the path difference and equating it to (n)(lambda)/2, but runs into an algebraic issue. Another member points out the error and provides a correction. The conversation then moves on to discussing the solution for part c of the problem, which the author struggles to understand.
  • #1
rteng
26
0

Homework Statement



1670a.jpg

1670b.jpg

1670c.jpg


Homework Equations



destructive interference occurs at 0.5(lambda), (3/2)(lambda),...



The Attempt at a Solution



I found the path difference to be:
d=sqrt(4+x^2)-x

and this has to be equal to (n)(lambda)/2 where n is an odd integer for destructive interference

this does not work out algebraically for me...as the x^2-x^2=0

is this path difference not correct?

do I just use f(n)=(nv)/(2d) to find d?
 
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  • #2
Why did you do: x^2-x^2=0? Where does that come from?
 
  • #3
ok I found my path difference to be sqrt(4+x^2)-x

so if I equate this to (n)(lambda)/2

then: sqrt(4+x^2)-x=(n)(lambda)/2

4+x^2-x^2=(n^2)(lambda^2)/4

there is my problem...
 
  • #4
Do you know how to find lambda?
 
  • #5
Oh, (sqrt(4 + x^2) - x)^2 = (sqrt(4 + x^2) - x) * (sqrt(4 + x^2) - x)
You don't just square the terms.
 
  • #6
ah yes...of course

I think that should be the proper way to solve this problem...at least for a and b

now on to c...
 
  • #7
ok I do not understand c
can anybody help?
 
  • #8
I would say that the only way for which there be no destructive interference is if x is 0
but that revelation is really vague for me...could it be correct?
 
  • #9
the I get f=86Hz
 

1. What is destructive interference?

Destructive interference is a phenomenon that occurs when two waves with the same frequency and amplitude meet and cancel each other out, resulting in a decrease in the overall amplitude of the resulting wave. This can occur when the peaks of one wave align with the troughs of the other wave.

2. How does destructive interference affect sound?

In sound, destructive interference can result in a decrease in the volume or loudness of the sound. This can occur when two sound waves with the same frequency and amplitude meet and cancel each other out, resulting in a quieter sound or even complete silence.

3. What are some real-life examples of destructive interference?

One example of destructive interference is when two identical sound waves from speakers collide and cancel each other out, resulting in a decrease in volume. Another example is when waves in a body of water, such as waves from two boats, meet and cancel each other out, resulting in calmer waters.

4. How does the distance between waves affect destructive interference?

The distance between waves can affect destructive interference by changing the phase difference between the waves. If the waves are in phase (peaks and troughs align), destructive interference will occur. However, if the waves are out of phase (peaks and troughs do not align), constructive interference will occur.

5. Can destructive interference be used beneficially?

Yes, destructive interference can be used beneficially in a process called noise-cancelling. This involves generating a sound wave that is out of phase with the unwanted sound, effectively cancelling it out and reducing noise pollution. Destructive interference is also used in certain musical instruments, such as string instruments, to produce specific tones and harmonics.

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