Destructive interference and constructive interference

  • Thread starter shawen
  • Start date
  • #1
3
0
Moved from general physics forum
Two loudspeakers are placed facing each 4.0 m apart. The speakers emit sound in phase with a frequency of 266 Hz. An observer at the center (2.0 m from each speaker along the line joining them) experiences constructive interference. The speed of sound is 343 m/s

The lowest frequency at which destructive interference could occur is if 1/2 a wavelength = 1.5 m (the difference between the two)

The speed of sound can be found with the following formula
v = 331 m/s + (0.6 m/s/C)•T
So at 20C you get a speed of 343 m/s.

We know that (1/2)*lambda = 1.5 m, so lambda = 3.0m
The wavelength must be 3 m.

Using the universal wave equation

v = f*lambda
f = v/lambda
f = (343m/s)/(3 m)
= 114 Hz

The lowest freq would be 114 Hz

The next two lowest would be a 3lambda/2 = 1.5 and 5lambda/2 = 1.5

so lambda = 1m and lambda = 0.6m

Those give f = 343 Hz and 572 Hz

i don't know how to find How far toward either speaker must the observer walk to experience destructive interference?
please help me
 
Last edited:

Answers and Replies

  • #2
jtbell
Mentor
15,716
3,845
i don't know how to find how far it is
How far what is? (What is "it"?)
 
  • #3
3
0
How far what is? (What is "it"?)
How far toward either speaker must the observer walk to experience destructive interference?
 
  • #4
nasu
Gold Member
3,769
428
What is the c
How far toward either speaker must the observer walk to experience destructive interference?
What is the condition for destructive interference? What path difference will give destructive interference?
 

Related Threads on Destructive interference and constructive interference

Replies
5
Views
4K
Replies
6
Views
1K
Replies
26
Views
24K
Replies
17
Views
1K
  • Last Post
Replies
5
Views
9K
Replies
5
Views
930
Replies
6
Views
2K
Replies
0
Views
2K
Replies
3
Views
48K
Replies
4
Views
5K
Top