Destructive interference and constructive interference

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  • #1
shawen
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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
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i don't know how to find how far it is

How far what is? (What is "it"?)
 
  • #3
shawen
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How far what is? (What is "it"?)
How far toward either speaker must the observer walk to experience destructive interference?
 
  • #4
nasu
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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?
 

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