• Support PF! Buy your school textbooks, materials and every day products Here!

Can you please check this (destructive interference).

  • Thread starter Sullo
  • Start date
  • #1
15
0
If it is wrong can you please pinpoint where i am going wrong. Thank you.

1. Homework Statement

https://imgur.com/a/0STmmWt
upload_2018-5-19_15-17-56.png

(uploaded picture because it has a diagram)

Homework Equations


To sense destructive interference , the person must be one half wavelength (0.5λ) closer or farther from one speaker than the other.

The Attempt at a Solution


Choosing distance from closer speaker (speaker on the right). So to sense destructive frequency he must be 0.5λ closer to the right speaker than the left.
Distance from right speaker before moving to the right (c^2) = a^2 + b^2
Distance from right speaker before moving to the right (c^2) = 1.75^2 + 5.0^2
Distance from right speaker before moving c = 5.30m

Distance from right speaker after moving to the right (c^2) = a^2 + b^2
Distance from right speaker after moving to the right c^2 = (1.75-0.84)^2 + (5.0)^2
Distance from right speaker after moving to the right c = 5.08m

Therefore; Distance from right speaker before moving - Distance from right speaker after moving
=5.30 - 5.08 = 0.22m

Therefore ; 0.22m = 0.5λ --> λ = 0.44m

v = fλ , f = v/λ = 343/0.44 = 779.5Hz
 

Attachments

Last edited by a moderator:

Answers and Replies

  • #2
34,074
9,983
The distance to the left speaker changes as well.
 
  • #3
15
0
The distance to the left speaker changes as well.
Can you please elaborate, why include both speakers in this?
 
  • #4
34,074
9,983
To sense destructive interference , the person must be one half wavelength (0.5λ) closer or farther from one speaker than the other.
The length difference between the left and right speaker has to be half the wavelength. What you calculated is something else - you didn't even consider the left speaker.
 
  • #5
15
0
The length difference between the left and right speaker has to be half the wavelength. What you calculated is something else - you didn't even consider the left speaker.
I'll need to go over this again tomorrow then
 
  • #6
15
0
The length difference between the left and right speaker has to be half the wavelength. What you calculated is something else - you didn't even consider the left speaker.
So do you mean distance from left speaker after moving + distance from right speaker after moving = 0.5λ ?
 
  • #7
34,074
9,983
The difference, not the sum.
 
  • #8
15
0
The difference, not the sum.
Oh ok. Thank you!

So is this correct?

Distance from left speaker after moving to the right (c^2) = a^2 + b^2
Distance from left speaker after moving to the right (c^2) = (1.75+0.84)^2 + 5.0^2
Distance from left speaker after moving to right c = 5.631

Distance from right speaker after moving to the right (c^2) = a^2 + b^2
Distance from right speaker after moving to the right c^2 = (1.75-0.84)^2 + (5.0)^2
Distance from right speaker after moving to the right c = 5.08m

Therefore; Distance from left speaker after moving - Distance from right speaker after moving
=5.631 - 5.08 = 0.551m

Therefore ; 0.551m = 0.5λ --> λ = 1.102m

v = fλ , f = v/λ = 343/1.102 = 311.26 Hz , 310hz (2 sig figs).
 
  • #9
34,074
9,983
That looks good.
 

Related Threads on Can you please check this (destructive interference).

Replies
7
Views
1K
Replies
1
Views
467
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
1
Views
998
  • Last Post
Replies
5
Views
1K
Replies
6
Views
1K
Replies
1
Views
4K
Replies
1
Views
1K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
4
Views
12K
Top