Question

Two loudspeakers are placed on a wall 2 m apart. A listener stands directly in front of one of the speakers, 81.7 m from the wall. The speakers are being driven by the same electric signal generated by a harmonic oscillator of frequency 4450 Hz. What is the phase difference ∆Φ between the two waves (generated by each speaker) when they reach the listener? The speed of the sound in air is 343 m/s. Answer in units of rad.

Answer 1

Answer:

The phase difference is       $\Delta \phi = 1.9995 rad$  

Explanation:

From the question we are told that

    The distance between the  loudspeakers is $d = 2m$

     The distance of the listener from the wall  $D = 81.7 \ m$

     The frequency of the  loudspeakers is  $f = 4450Hz$

      The velocity of sound is $v_s = 343 m/s$

     

The path difference of the sound wave that is getting to the listener is mathematically represented as

        $\Delta z =\sqrt{d^2 + D^2} -D$

Substituting values

        $\Delta z =\sqrt{2^2 + 81.7^2 } -81.7$

       $\Delta z =0.0245m$

The phase difference is mathematically represented as

           $\Delta \phi$ =  $\frac{2 \pi}{\lambda } * \Delta z$

Where $\lambda$ is the wavelength which is mathematically represented as

          $\lambda = \frac{v_s }{f}$

substituting value  

          $\lambda = \frac{343 }{4450}$

        $\lambda = 0.0770 m$

Substituting value into the  equation for phase difference

      $\Delta \phi$ =  $\frac{2 * 3.142 * 0.0245}{0.0770}$

      $\Delta \phi = 1.9995 rad$