# Sound source A is located at x = 0, y = 0, and sound source B is place at x = 0, y = 2.7 m. The...

## Question:

Sound source {eq}A {/eq} is located at {eq}x = 0,\ y = 0 {/eq} , and sound source {eq}B {/eq} is place at {eq}x = 0,\ y = 2.7 \ m {/eq}. The two sources radiate coherently in phase. An observer at {eq}x = 30 \ m, \ y = 0 {/eq} notes that as she takes a few steps in either the positive or negative {eq}y {/eq} direction away from {eq}y = 0 {/eq} , the sound intensity diminishes.

a. What is the lowest frequency of the sources that can account for that observation? Answer in units of {eq}Hz {/eq}.

b. What is the next higher frequency of the sources that can account for that observation? Answer in units of {eq}Hz {/eq}.

## Interference of Sound Waves:

If two sound sources are emitting sound in phase and the frequency of two waves are the same, the resulting intensity distribution in space will form so called interference pattern, so that intensity is at maximum at certain points and is nearly zero at others. The condition for the constructive interference is such that the path length difference between two waves reaching the point of observation equals the integer number of wavelengths of the wave.

a) The point of observation corresponds to the point of constructive interference of sound. The path-length difference between two waves is then given by the following equation:

{eq}\sqrt {x^2_1 + y^2_1} - x_1 = m \lambda {/eq}

Here

• {eq}x_1 = 30 \ m {/eq} is the x-coordinate of the point of observation and the distance to the source at the origin;
• {eq}y_1 = 2.7 \ m {/eq} is the y-coordinate of the second source;
• {eq}m {/eq} is the order of maximum;
• {eq}\lambda = \dfrac {v}{f} {/eq}

is the wavelength of the sound wave;

• {eq}v = 343 \ m/s {/eq} is the speed of sound in the air;

The lowest frequency corresponds to the lowest order maximum, therefore,

{eq}m = 1 {/eq}.

Solving for the lowest frequency, we obtain:

{eq}f = \dfrac {v}{\sqrt{x^2_1 + y^2_1} - x_1} {/eq}

Calculating, we get:

{eq}f = \dfrac {343 \ m/s}{\sqrt{(30 \ m)^2 + (2.7 \ m)^2} - 30 \ m} \approx \boxed {\color{blue}{2829 \ Hz}} {/eq}

b) The second-highest frequency corresponds to the second-order maximum (m = 2)

Then we obtain:

{eq}f = \dfrac {2v}{\sqrt{x^2_1 + y^2_1} - x_1} {/eq}

Calculating, we get:

{eq}f = \dfrac {2\cdot 343 \ m/s}{\sqrt{(30 \ m)^2 + (2.7 \ m)^2} - 30 \ m} \approx \boxed {\color{red}{5658 \ Hz}} {/eq}