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Sound frequency from distance b/w first minimum and maximum

  1. Jul 21, 2016 #1
    1. The problem statement, all variables and given/known data
    Two speakers are arranged as shown. For this problem, assume that point O is 12 m along the center line and the speakers are separated by a distance of 1.5 m. As the listener moves toward point P from point O, a series of alternating minima and maxima is encountered. The distance between the first minimum and the next maximum is 0.4 m. Using 344 m/s as the speed of sound in air, determine the frequency of the speakers. (Use the approximation of sin(Θ)=tan(Θ).

    Speakers_zpsksnibpl8.jpg

    2. Relevant equations
    ƒ = v/λ
    r1 = distance from top speaker to observer
    r2 = distance from bottom speaker to observer

    3. The attempt at a solution
    Assumptions:
    Δr = λ/2 for first minimum (corresponding to phase shift of π)
    Δr = λ for first maximum after first minimum (corresponding to phase shift of 2π)

    I guess the trig/algebra is tripping me up here. The question specifically calls out that sin can be approximated as tan. I'm not sure how to use this, what I assume to be, useful information. I think I need to find the location of the first minimum before I can solve, but I'm not sure. I've tried to determine this using pythagorean ratios, but I keep coming up with lousy answers. Would anyone be willing to point me in the right direction on how to get through the first step of this? If the problem can be solved without finding the location of the first minimum, let me know that as well please.
     
    Last edited: Jul 21, 2016
  2. jcsd
  3. Jul 21, 2016 #2

    collinsmark

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    Homework Helper
    Gold Member

    In terms of the [itex] \sin \theta \approx \tan \theta [/itex], this is a valid approximation for small [itex] \theta [/itex].

    To realize how it might help you in this problem, first review the following:

    [itex] \sin \theta = \frac{\mathrm{opposite}}{\mathrm{hypotenuse}} [/itex]

    [itex] \cos \theta = \frac{\mathrm{adjacent}}{\mathrm{hypotenuse}} [/itex]

    [itex] \tan \theta = \frac{\mathrm{opposite}}{\mathrm{adjacent}} [/itex]

    The approximation helps [might help] you because if you want to calculate the sine of an angle, you might have to calculate the hypotenuse all the time using the ugly [itex] c = \sqrt{a^2 + b^2} [/itex], but if instead you use the tangent approximation, you can substitute the hypotenuse with the adjacent. You already know what the adjacent is: it's 12 m for all your calculations. :wink: (one part of the figure indicates it's 8 m, but I think that's a mistake in the figure.)

    [Edit: although thinking about it a little more, I'm not completely sure if that approximation would help with this particular problem without making one or two other additional approximations. Hmm.]
     
    Last edited: Jul 22, 2016
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