# Two sound sources 1 and 2 and point P are located on an isosceles, right triangle with d = 0.50...

## Question:

Two sound sources 1 and 2 and point P are located on an isosceles, right triangle with d = 0.50 m. The sources are in phase and they both emit a sound of frequency 100 Hz. The sound propagates through air.

What is the phase difference between the two waves at point P?

## Superposition Of Sound waves

When two sound waves superpose, a new wave is formed and phenomenon is called interference. There are some places having maximum loudness(constructive interference) and some having minimum loudness(destructive interference).

We can get the phase difference between the interfering waves, getting the path difference.

Path difference and phase difference are {eq}\displaystyle \Delta x , \Delta \phi {/eq}

Formula is {eq}\Delta \phi = \dfrac{2\pi}{\lambda} \Delta x {/eq}

The path difference between two waves reaching point P {eq}\Delta x = 0.71 - 0.50 = 0.21 m {/eq}

We can use the formula {eq}v = f \lambda {/eq} v is velocity and " f " is frequency of sound, here we have considered v = 330 m/s

{eq}\lambda = \dfrac{330}{100} = 3.33 m {/eq}

The phase difference {eq}\Delta \phi = \dfrac{2\pi}{3.33} 0.21 \\ \Delta \phi =0.40 \,\mathrm{radian} {/eq} 