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
[tex]\Delta x = n \lambda \rightarrow \lambda = \frac{\Delta x}{n}[/tex]
Where
[tex]\Delta x =[/tex] Distance between speakers
n = Integer which represent he number of repetition of the spectrum
[tex]\lambda =[/tex] Wavelength
At the same time the frequency is subject to the form,
[tex]f = \frac{v}{\lambda}[/tex]
Where
[tex]v = 343m/s \rightarrow[/tex] Velocity (of the sound at this case)
From our given values we have to [tex]\Delta x[/tex] is
[tex]\Delta x = 50.5m-26m[/tex]
[tex]\Delta x = 24.5m[/tex]
The wavelength would be subject to the sound spectrum therefore for n = 1,
[tex]\lambda = \frac{\Delta x}{n}[/tex]
[tex]\lambda = \frac{24.5}{1}[/tex]
[tex]\lambda = 24.5m[/tex]
Then the frequency would be,
[tex]f = \frac{v}{\lambda}[/tex]
[tex]f = \frac{343}{24.5}[/tex]
[tex]f = 14Hz[/tex]
For the value of n = 2,
[tex]\lambda = \frac{\Delta x}{n}[/tex]
[tex]\lambda = \frac{24.5}{2}[/tex]
[tex]\lambda = 12.25m[/tex]
Then the frequency would be,
[tex]f = \frac{v}{\lambda}[/tex]
[tex]f = \frac{343}{12.25}[/tex]
[tex]f = 28Hz[/tex]
For n = 3,
[tex]\lambda = \frac{\Delta x}{n}[/tex]
[tex]\lambda = \frac{24.5}{3}[/tex]
[tex]\lambda = 8.167m[/tex]
Then the frequency would be,
[tex]f = \frac{v}{\lambda}[/tex]
[tex]f = \frac{343}{12.25}[/tex]
[tex]f = 42Hz[/tex]
From the frequencies obtained we can identify that the two lowest frequencies that can be heard due to constructive interference are 28Hz and 42Hz
