Question

Suppose that the separation between two speakers A and B is 4.30 m and the speakers are vibrating in-phase. They are playing identical 103-Hz tones and the speed of sound is 343 m/s. An observer is seated at a position directly facing speaker B in such a way that his line of sight extending to B is perpendicular to the imaginary line between A and B. What is the largest possible distance between speaker B and the observer, such that he observes destructive interference

Answer 1

Answer:

The largest  possible distance is $x = 4.720 \ m$

Explanation:

From the question we are told that

    The distance of  separation is   $d = 4.30 \ m$

      The  frequency of the tone played by both speakers is $f = 103 \ Hz$

     The speed of sound is  $v_s = 343 \ m/s$

The  wavelength of the tone played by the speaker is  mathematically evaluated as

              $\lambda = \frac{v}{f}$

substituting values

            $\lambda = \frac{343}{103}$

            $\lambda = 3.33 \ m$

Let the the position of the observer be O

Given that the line of sight between observer and speaker B is  perpendicular to the distance between A and B then

        The distance between A and the observer is  mathematically evaluated using Pythagoras theorem as follows

               $L = \sqrt{d^2 + x^2}$

Where x is the distance between the observer and B

  For the observer to observe destructive interference

          $L - x = \frac{\lambda}{2}$

So  

          $\sqrt{d^2 + x^2} - x = \frac{\lambda}{2}$

       $\sqrt{d^2 + x^2} = \frac{\lambda}{2} +x$

        $d^2 + x^2 = [\frac{\lambda}{2} +x]^2$

         $d^2 + x^2 = [\frac{\lambda^2}{4} +2 * x * \frac{\lambda}{2} + x^2]$

       $d^2 = [\frac{\lambda^2}{4} +2 * x * \frac{\lambda}{2} ]$

substituting values              

       $4.30^2 = [\frac{3.33^2}{4} +2 * x * \frac{3.33}{2} ]$

      $x = 4.720 \ m$