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 speak

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.

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Xavia 6 months 2021-07-20T12:14:11+00:00 1 Answers 11 views 0

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    2021-07-20T12:15:14+00:00

    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  

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