1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Speakers & sound waves

  1. Nov 10, 2007 #1
    Two loudspeakers are placed side by side a distance d apart. A listener observes constructive interference while standing in front of the loudspeakers, equidistant from both of them. The distance from the listener to the point half-way between the speakers is l.

    One of the loudspeakers is then moved directly away from the other. Once the speaker is moved a distance r from its original position, the listener, who is not moving, observes destructive interference for the first time.

    Find the speed of sound v in the air if both speakers emit a tone of the same frequency f.
    http://session.masteringphysics.com/problemAsset/1000054/11/104507C.jpg
    ______________________________

    I know that velocity = (wavelength)(frequency), and the path length difference for the case of destructive interference is =0.5(wavelength). And the distance between the observer and the speaker that has been moved is sqrt((0.5d+r)^2 + l^2).

    How do I put everything together to get the speed of sound in the air using only those variables introduced in the problem????
     
  2. jcsd
  3. Nov 10, 2007 #2
    What is the distance between the observer and the speaker that has not been moved?

    What is the difference between the two distances?
     
  4. Nov 10, 2007 #3
    the distance between the observer and the speaker that has not been moved is sqrt(0.5d^2+l^2) and the difference is sqrt((0.5d+r)^2 + l^2)-sqrt(0.5d^2+l^2)

    now what????

    I already had these in my head, but I didn't know how to follow through and put it together. Anyone can help me?
     
    Last edited: Nov 10, 2007
  5. Nov 10, 2007 #4
    So far so good. That's the difference expressed in terms of d, l and r. What is it expressed in terms of wavelength?
     
  6. Nov 10, 2007 #5
    I don't know. Every wavelength cycle measures 2pi so (wavelength)(sqrt((0.5d+r)^2 + l^2)-sqrt(0.5d^2+l^2)) / 2pi . Am I right?
     
  7. Nov 11, 2007 #6
    You do know! "I know that ... the path length difference for the case of destructive interference is =0.5(wavelength). "
     
  8. Nov 11, 2007 #7
    You know, the simplest way to do this question is to make use of the young's double slit experiment equations? This is effectively the same thing, only the waves are sound waves and the screen is the person.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?