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

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

Step-by-step 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 [(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 [(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 [( a)/(\lambda) + ft ] - 2 \pi [(b)/(\lambda) + ft ]`

=>
`\Delta \phi = (2\pi [ a - b])/( \lambda )`

Gnerally the wavelength is mathematically represented as

`\lambda = (v)/(f)`

=>
`\lambda = (330)/(540)`

=>
`\lambda = 0.611 \ m`

=>
`\Delta \phi = (2* 3.142 [ 4.40 - 4.0 ])/( 0.611 )`

=>
`\Delta \phi = 4.12 \ rad`