Answer:
Approximately [tex]30.0\; \rm Hz[/tex].
Explanation:
Look up the speed of sounds in the air at [tex]25\; ^\circ\rm C[/tex]: [tex]v \approx 346\; \rm m \cdot s^{-1}[/tex].
Let the frequency of this tune be [tex]f\; \rm Hz[/tex]. The wavelength of the tune would be [tex]\displaystyle \lambda = \frac{v}{f} \approx \frac{346}{f}[/tex].
The distance between the first speaker and the listener is [tex]12.608\; \rm m[/tex]. How many wavelengths can fit into that distance?
[tex]\displaystyle \frac{12.608}{\lambda} \approx \frac{12.608}{346 / f} = \frac{12.608}{346} \cdot f[/tex].
Similarly, the distance between the second speaker and the listener is [tex]18.368\; \rm m[/tex]. Number of wavelengths in that distance:
[tex]\displaystyle \frac{18.368}{\lambda} \approx \frac{18.368}{346 / f} = \frac{18.368}{346} \cdot f[/tex].
Difference between these two numbers:
[tex]\begin{aligned} &\frac{18.368}{346} \cdot f - \frac{12.608}{346} \cdot f = \frac{5.76}{346}\cdot f\end{aligned}[/tex].
For destructive interference to occur, that difference should be equal to [tex]\displaystyle \frac{1}{2}[/tex], [tex]\displaystyle 1 + \frac{1}{2}= \frac{3}{2}[/tex], [tex]\cdots[/tex], or [tex]\displaystyle \left(k+ \frac{1}{2}\right)[/tex] in general ([tex]k[/tex] can be any non-negative whole number.)
Let [tex]\begin{aligned} \frac{5.76}{346}\cdot f = k + \frac{1}{2}\end{aligned}[/tex], and solve for [tex]f[/tex].
[tex]\begin{aligned} f&= \left.\left(k + \frac{1}{2}\right) \right/\frac{5.76}{346}\\ &= \frac{346}{5.76} \, k + \frac{1}{2} \times \frac{346}{5.76} \\ &\approx 60.1\, k + 30.0 \end{aligned}[/tex].
The next step is to find the values of [tex]k[/tex] that ensure [tex]20 \le f \le 20\times 10^{3}[/tex] (frequency is between [tex]20\; \rm Hz[/tex] and [tex]20\, \rm kHz[/tex].) It turns out that [tex]k = 0[/tex] (the smallest [tex]k[/tex] value possible) would be sufficient. In that case, the frequency is approximately [tex]30.0\; \rm Hz[/tex]. Using a larger [tex]k[/tex] would only increase the frequency.
