Finding closest distance that will generate destructive interference

In summary: Yes, the equation for d2 is d2=d1-wavelength(n-1/2). The equation for d1 is sqrt(4.6^2+d^2). When you solve for n, the closest integer value to 1.24 is 1.
  • #1
JoeyBob
256
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Homework Statement
See attached
Relevant Equations
d1-d2=wavelength(n-1/2)
So I can calculate wavelenghth using v/f to get 6.24. Now I made d1=sqrt(4.6^2+d^2) and d2=d so my equation now looks like

sqrt(4.6^2+d^2)-d =6.24(n-1/2). To find n I made d=0

4.6=6.24(n-1/2) n=1.24

Now my problem is that I don't know which n to use or if I've even done the above correctly. It could be 0.24(wavelength), 0.74(wavelength), 1.24(wavelength) ect.

Furthermore the correct answer is d=1.83 which is indicative of n=1 which isn't what i calculated.
 

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  • #2
JoeyBob said:
Homework Statement:: See attached
Relevant Equations:: d1-d2=wavelength(n-1/2)

So I can calculate wavelenghth using v/f to get 6.24. Now I made d1=sqrt(4.6^2+d^2) and d2=d so my equation now looks like

sqrt(4.6^2+d^2)-d =6.24(n-1/2). To find n I made d=0
You cannot make d = 0 because d is what you are looking for. If you meant to say n instead of d, you cannot set n=0 either because that will make the right hand side of the equation negative wile the left hand side is positive for any n. What does this suggest about the value of n? Considering the relevant equation you wrote, what are the possible values of n?
 
  • #3
kuruman said:
You cannot make d = 0 because d is what you are looking for. If you meant to say n instead of d, you cannot set n=0 either because that will make the right hand side of the equation negative wile the left hand side is positive for any n. What does this suggest about the value of n? Considering the relevant equation you wrote, what are the possible values of n?

So 1, 2, 3, 4, 5, ect. could be n? But how would I know which one to pick besides trying them out all individually to find the smallest d?
 
  • #4
Read the problem. It's asking for the closest location. What does that mean in terms of the value of n?

On edit: Suppose you start at d = 0 and you move out increasing d. Which of the two are you going to encounter first, a maximum or a minimum and why?
 
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  • #5
kuruman said:
Read the problem. It's asking for the closest location. What does that mean in terms of the value of n?

On edit: Suppose you start at d = 0 and you move out increasing d. Which of the two are you going to encounter first, a maximum or a minimum and why?

Wait so did I have the right idea? I made d=0, so n=1.24, which is closest to 1?

Does n have to be an integer?
 
  • #6
JoeyBob said:
Wait so did I have the right idea? I made d=0, so n=1.24, which is closest to 1?
Does n have to be an integer?
You have the right equation, which is d1-d2=wavelength(n-1/2). It seems to me that you are not sure what it means. So study your notes or textbook, visit the web if you have to and explain to me what each of the symbols means in terms of the problem that you are asked to solve. I will provide a template for you, but you have to fill in the blanks.

Variables
d1 = the distance from _________ to _________ . In this problem it can be written as _________ .
d2 = the distance from _________ to _________ . In this problem it can be written as _________ .
wavelength = the wavelength of the sound. In this problem it can be written as λ = 6.24 m (you got that right).
n = _____________________ .

Equation interpretation
Use your own words to describe what this equation means and when it is applicable.

Yes n has to be an integer and when you understand interference, you will understand why. The value you got n = 1.24 is not an integer equal to 1 ever for small values of 1.24.
 
  • #7
kuruman said:
Yes n has to be an integer and when you understand interference, you will understand why. The value you got n = 1.24 is not an integer equal to 1 ever for small values of 1.24.

Yes but the closest integer value to 1.24 is 1. I was wondering if this procedure would get me the right answer for these sorts of porblems, where I make d=0, solve for n, then find the integer that is closest to the n value I found.

In this case I made d=0, solved for n, found n=1.24 for d=0, and now I'm wondering if using the closest integer value will consistently get me the correct values for these sorts of problems. For this specific problem it obviously does.

I believe the fact that I applied the correct equations for d2 and d1 means that I know what they mean, as well as the wavelength. My question right now is whether my above solution procedure will give correct answers consistently or if its only a fluke that works for this specific problem.
 
  • #8
Your method works for this particular problem. I would not recommend that you follow it blindly. If you wish to test your understanding for this sort of problem, consider the following variation:

The speaker separation is increased from to more than twice its previous value keeping the sound frequency the same. Destructive interference is observed at d = 5.5 m from the left speaker. Find the new speaker separation if the set up fits in a 15×15 m2 square room.

You don't have to solve this problem if you don't want to. However, if you do and show me your solution, I will show you mine and discuss it with you.

Edit: The new separation is specified to be "more than twice its previous value."
 
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  • #9
kuruman said:
Your method works for this particular problem. I would not recommend that you follow it blindly. If you wish to test your understanding for this sort of problem, consider the following variation:

The speaker separation is increased from its previous value keeping the sound frequency the same. Destructive interference is observed at d = 5.5 m from the left speaker. Find the new speaker separation if the set up fits in a 15×15 m2 square room.

You don't have to solve this problem if you don't want to. However, if you do and show me your solution, I will show you mine and discuss it with you.

So d2-d1=(n-1/2)wavelenghth then

sqrt(5.5^2+x^2)-5.5 = (n-1/2)wave

d=0 then

5.5=(n-1/2)wave. let's say wave=1

6.5=n

Now I just find the n value 1.5, 2.5, 3.5 ect that fits in such a room?
 
  • #10
First, let me say that I am pleased that you chose to pursue this further. Having said that, I will add that this is your solution based on your understanding of interference. Follow it through and get a number. Then, as promised, I will show you my solution and compare methods and numbers.
 

1. How is the closest distance determined for destructive interference?

The closest distance for destructive interference is determined by calculating the wavelength of the wave and dividing it by two. This results in the distance where the crests and troughs of the waves are perfectly aligned and cancel each other out, creating destructive interference.

2. What factors affect the closest distance for destructive interference?

The closest distance for destructive interference can be affected by the wavelength of the wave, the amplitude of the wave, and the distance between the two sources of the wave. Additionally, the medium through which the wave is traveling can also impact the closest distance for destructive interference.

3. How does changing the frequency of the wave affect the closest distance for destructive interference?

Changing the frequency of the wave does not directly affect the closest distance for destructive interference. However, it can impact the wavelength of the wave, which in turn can affect the closest distance. Higher frequencies have shorter wavelengths, resulting in a smaller closest distance for destructive interference.

4. Can the closest distance for destructive interference be negative?

No, the closest distance for destructive interference cannot be negative. It is always a positive value representing the distance between the two sources of the wave where destructive interference occurs. If the distance between the sources is less than the closest distance, constructive interference will occur instead.

5. How is the closest distance for destructive interference used in real-world applications?

The concept of the closest distance for destructive interference is used in various real-world applications, such as noise-cancelling headphones and acoustic panels. By placing two sources of sound at the closest distance for destructive interference, the unwanted sound waves can be cancelled out, resulting in a quieter environment.

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