Find smallest distance x from y-axis for both constructive & destructive interference to occur

  • Thread starter Thread starter hraghav
  • Start date Start date
  • Tags Tags
    Interference
Click For Summary
The discussion focuses on calculating the smallest distance from the y-axis for both constructive and destructive interference. For constructive interference, the calculated distance is approximately 99.90 m, while for destructive interference, it is around 49.76 m. However, both answers are questioned for accuracy, with suggestions to consider x = 0 as a possible solution. The calculations appear correct, but the final answers may require adjustments for significant figures. Participants emphasize the importance of entering values with the correct number of significant figures to obtain the right results.
hraghav
Messages
48
Reaction score
5
Homework Statement
Two speakers are located a distance 2d apart along the x-axis, where the origin is located at the midpoint between the two speakers. This is shown in the image below. The speakers emit sound with wavelength λ=1.34m. The distance from the origin to either of the speakers is d=6.74m. At distances y≫d, the lines connecting the speakers to the listener can be treated as effectively parallel, similar to two slit diffraction. In such a case, the difference in distance from the speakers to the listening point can be approximated as Δr=2dsinθ, where the angle θ is shown in green in the image above. For the following questions, assume that the listener is at a location
r→=(x,1000m).

a) What is the smallest distance x from the y-axis such that constructive interference occurs?
b) What is the smallest distance x from the y-axis such that destructive interference occurs?
Relevant Equations
For constructive interference: Δr=mλ
For destructive interference: Δr=(m+ 0.5)λ
a) For constructive interference: Δr=mλ where m = 1 for smallest distance
Δr= λ where Δr= 2dsinθ given in the question
so 2dsinθ = λ
θ = sin^-1( λ / 2d) where λ = 1.34m and 2d = 2*6.74 = 13.48m
θ = 5.7049 degrees
tan θ = x/1000
1000*tan(5.7049) = x
x = 99.9013 m

b) For destructive interference: Δr=(m+ 0.5)λ where m = 0 for smallest distance
Δr = 0.5 λ but Δr= 2dsinθ given in the question
2dsinθ = 0.5 λ
sinθ = λ / 4d where λ = 1.34m and 4d = 4*6.74 = 26.96m
θ = 2.8489 degrees

tan θ = x/1000
1000*tan(2.8489) = x
x = 49.76 m

Both these answers are not correct. Could someone please look at these and let me know where is the error?

Thank you

Screenshot 2024-07-25 at 5.18.07 PM.png


Screenshot 2024-07-25 at 5.19.07 PM.png
 
Physics news on Phys.org
Your work looks correct to me for the case where the listener is depicted by the green ear in the diagram. Do you know what answers you are supposed to get? Did you consider ##x = 0## as a possible answer?
 
TSny said:
Your work looks correct to me for the case where the listener is depicted by the green ear in the diagram. Do you know what answers you are supposed to get? Did you consider ##x = 0## as a possible answer?
0 worked for part a ie constructive interference but I still don't have the answer for part b. I can't see the final answers until and unless I don't enter the correct ones.
 
hraghav said:
0 worked for part a ie constructive interference but I still don't have the answer for part b. I can't see the final answers until and unless I don't enter the correct ones.
I still don't see any error in your calculation for (b).
 
TSny said:
I still don't see any error in your calculation for (b).
Sounds good thank you
 
hraghav said:
... but I still don't have the answer for part b. I can't see the final answers until and unless I don't enter the correct ones.
The answer may need to have an appropriate number of significant figures. It depends on the software checking the value.

Four significant figures is excessive. You could try entering the value rounded to 3 sig. figs.
 
Thread 'Chain falling out of a horizontal tube onto a table'
My attempt: Initial total M.E = PE of hanging part + PE of part of chain in the tube. I've considered the table as to be at zero of PE. PE of hanging part = ##\frac{1}{2} \frac{m}{l}gh^{2}##. PE of part in the tube = ##\frac{m}{l}(l - h)gh##. Final ME = ##\frac{1}{2}\frac{m}{l}gh^{2}## + ##\frac{1}{2}\frac{m}{l}hv^{2}##. Since Initial ME = Final ME. Therefore, ##\frac{1}{2}\frac{m}{l}hv^{2}## = ##\frac{m}{l}(l-h)gh##. Solving this gives: ## v = \sqrt{2g(l-h)}##. But the answer in the book...

Similar threads

  • · Replies 3 ·
Replies
3
Views
655
Replies
3
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
4K
  • · Replies 7 ·
Replies
7
Views
15K
  • · Replies 13 ·
Replies
13
Views
5K
  • · Replies 23 ·
Replies
23
Views
3K
  • · Replies 9 ·
Replies
9
Views
4K
Replies
1
Views
1K
Replies
20
Views
5K