# Two loudspeakers facing each other play identical sounds of the same frequency. Starting directly...

## Question:

Two loudspeakers facing each other play identical sounds of the same frequency. Starting directly between them, you begin to walk towards one of the speakers. As you do, the intensity of the sound you hear drops, then rises, then drops again to a second minimum. At this point, you measure your distance to the speakers and find that you are 5 m away from one speaker, and 1 m away from the other speaker. What is the frequency of the sound emitted by the speakers? Assume the speed of sound in air is 340 m/s.

## Interference of Waves:

Two waves of the same frequency that are coherent in space will interfere upon interaction, the amplitude being maximum at the points where the two waves are in phase, or the path length difference between the point of observation and the two sources of waves equals the integer number of wavelengths.

The difference in distances to the speakers in the second intensity minimum equals {eq}d_2 - d_1 = 1.5\lambda {/eq}

Here

• {eq}d_2 = 5 \ m {/eq} is the distance to the farthest speaker;
• {eq}d_1 = 1 \ m {/eq} is the distance to the closest speaker;
• {eq}\lambda {/eq} is the wavelength of the sound wave.

Calculating the wavelength, we obtain:

{eq}\lambda = \dfrac {5 \ m - 1 \ m}{1.5} \approx 2.67 \ m {/eq}

The frequency of the wave can be found from the equation, relating the wavelength, frequency, and the speed of the wave:

{eq}v = f\lambda {/eq}

Here

• {eq}v = 340 \ m/s {/eq} is the speed of sound in the air;

Solving for the frequency, we obtain:

{eq}f = \dfrac {v}{\lambda} {/eq}

Calculating, we get"

{eq}f = \dfrac {340 \ m/s}{2.67 \ m} = \boxed {127.5 \ Hz } {/eq} 