# Two small stereo speakers are 2.1 m apart and act as a double-slit interference experiment with...

## Question:

Two small stereo speakers are 2.1 m apart and act as a double-slit interference experiment with sound waves. If the opposite wall of the room is 6.0 m away and the sound frequency is 1.7 kHz, what is the distance between the two interference maxima nearest the center of the wall? Assume the speakers act as point sources of sound.

## Sound interference

The phenomenon of interference is a phenomenon of waves. Therefore, two point sources of sound waves can be treated the same as Young's double slit experiment. In this case it is only necessary to keep in mind that the wavelength of the sound waves must be calculated by the expression:

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

where {eq}v = 343\,\mathrm{m/s} {/eq} is the speed of sound in the air, and {eq}f {/eq} is the frequency of the wave.

As in Young's experiment, the maximums are equally spaced and their linear position on the screen can be calculated using the trigonometric ratio of the tangent:

{eq}\left.\begin{matrix} \lambda=\frac{v}{f}\\ d\sin\theta=m\lambda\\ \tan\theta=\frac{y}{D} \end{matrix}\right\}\Rightarrow \frac{\sin\theta}{\sqrt{1-\sin^2\theta}}=\frac{y}{D}\Rightarrow \frac{1}{\sqrt{\left ( \frac{df}{mv} \right )^2-1}}=\frac{y}{D}\\ y=\frac{D}{\sqrt{\left ( \frac{df}{mv} \right )^2-1}}=\mathrm{(6.0\,m)\left ( \left ( \frac{(2.1\,m)(1700\,Hz)}{(1)(343\,m/s)} \right )^2-1 \right )^{-1/2}=0.58\,m} {/eq}

The distance that separates the first order maxima (maximums that are next to the central maximum) from the center of the interference pattern is 58 cm. 