Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Frequency of first intensity maximum for loudspeakers

  1. Dec 13, 2008 #1
    1. The problem statement, all variables and given/known data
    Two loudspeakers are set up 4.0 meters apart and are driven in phase by the same amplifier. The listener is located a sufficient distance away so the angle to the listener is approximately the same at each speaker as indicated in the drawing. The speed of sound is 344 m/s. What frequency would the speakers need to emit if the first intensity maximum from the central maximum occurs at an angle of 30° from the perpendicular to each speaker?
    http://omploader.org/vMTBsaw/phys_diagram1.png [Broken]

    2. Relevant equations
    [tex]v=\lambda \cdot f[/tex]
    There is also [tex]A = 2 A_0 cos(\frac{\phi}{2})[/tex] in the solution to the problem but I don't understand where this equation even came from. It's not in my book or anything.

    3. The attempt at a solution
    Well, I know I'm looking for f, the frequency. So I did [tex]f = \frac{v_{\mbox{sound}}}{\lambda}[/tex] but that's about it. Now the actual solution uses [tex]A=2A_0 cos(\frac{\phi}{2})[/tex] and [tex]\phi = k \Delta x[/tex]. I would like to understand where these equations came from. Are they universal for this type of problem, or were they somehow derived from the given information?

    http://omploader.org/vMTBsbQ/phys_diagram2.png [Broken]
    Last edited by a moderator: May 3, 2017
  2. jcsd
  3. Dec 15, 2008 #2
    Surely the first max will occur when the path length from one speaker is a single wavelength longer than to the path length to the other.
    The path difference shown here is 4*sin30° = 2metres
    So 344m/sec = 2m * frequency
    Frequency = 172 Hertz.
    Last edited: Dec 15, 2008
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook