Question

Two loudspeakers on a concert stage are vibrating in phase. A listener 50.5 m from the left speaker and 26.0 m from the right one. the listener can respond to all frequencies from 20 to 20 000Hz, and the speed of sound is 343 m/s. What are the two lowest frequencies that can be heard loudly due to constructive interference?

Answer 1

To solve this problem, it is necessary to apply the concepts related to the constructive interference caused by the wavelengths of the sound traveling in the air.

From the definition of constructive interference we know that

$\Delta x = n \lambda \rightarrow \lambda = \frac{\Delta x}{n}$

Where

$\Delta x =$ Distance between speakers

n = Integer which represent he number of repetition of the spectrum

$\lambda =$ Wavelength

At the same time the frequency is subject to the form,

$f = \frac{v}{\lambda}$

Where

$v = 343m/s \rightarrow$ Velocity (of the sound at this case)

From our given values we have to $\Delta x$ is

$\Delta x = 50.5m-26m$

$\Delta x = 24.5m$

The wavelength would be subject to the sound spectrum therefore for n = 1,

$\lambda = \frac{\Delta x}{n}$

$\lambda = \frac{24.5}{1}$

$\lambda = 24.5m$

Then the frequency would be,

$f = \frac{v}{\lambda}$

$f = \frac{343}{24.5}$

$f = 14Hz$

For the value of n = 2,

$\lambda = \frac{\Delta x}{n}$

$\lambda = \frac{24.5}{2}$

$\lambda = 12.25m$

Then the frequency would be,

$f = \frac{v}{\lambda}$

$f = \frac{343}{12.25}$

$f = 28Hz$

For n = 3,

$\lambda = \frac{\Delta x}{n}$

$\lambda = \frac{24.5}{3}$

$\lambda = 8.167m$

Then the frequency would be,

$f = \frac{v}{\lambda}$

$f = \frac{343}{12.25}$

$f = 42Hz$

From the frequencies obtained we can identify that the two lowest frequencies that can be heard due to constructive interference are 28Hz and 42Hz