Question

Two sound waves, from two different sources with the same frequency, 540 Hz, travel in the same direction at 330 m s . The sources are in phase. What is the phase difference of the waves at a point that is 4.40 m from one source and 4.00 m from the other?

Answer 1

Answer:

The value is $\Delta \phi = 4.12 \ rad$

Explanation:

From the question we are told that

    The frequency of each sound is  $f_1 = f_2 = f = 540 \ Hz$

      The speed of the sounds is  $v = 330 \ m/s$

       The  distance of the first source from the point considered is  $a = 4.40 \ m$

        The distance of the second source from the point considered is  $b = 4.00 \ m$

Generally the phase angle made by the first sound wave at the considered point is mathematically represented as

           $\phi_a = 2 \pi [\frac{a}{\lambda} + ft]$

Generally the phase angle made by the first sound wave at the considered point is mathematically represented as

           $\phi_b = 2 \pi [\frac{b}{\lambda} + ft]$          

Here b is the distance o f the first wave from the considered point  

Gnerally the phase diffencence is mathematically represented as  

           $\Delta \phi= \phi_a - \phi_b = 2 \pi [\frac{ a}{\lambda} + ft ] - 2 \pi [\frac{b}{\lambda} + ft ]$      

=>      $\Delta \phi = \frac{2\pi [ a - b]}{ \lambda }$

Gnerally the wavelength is mathematically represented as

        $\lambda = \frac{v}{f}$

=>     $\lambda = \frac{330}{540}$

=>     $\lambda = 0.611 \ m$

=>    $\Delta \phi = \frac{2* 3.142 [ 4.40 - 4.0 ]}{ 0.611 }$

=>    $\Delta \phi = 4.12 \ rad$