Question

Two loudspeakers are located 4.965 m apart on an outdoor stage. A listener is 12.608 m from one and 18.368 m from the other. During the sound check, a signal generator drives the two speakers in phase with the same amplitude and frequency. The transmitted frequency is swept through the audible range (20 Hz to 20 kHz). What is the lowest frequency that gives minimum signal (destructive interference) at the listener’s location? The air temperature is 25.0 degrees Celsius.

Answer 1

Answer:

Approximately $30.0\; \rm Hz$.

Explanation:

Look up the speed of sounds in the air at $25\; ^\circ\rm C$: $v \approx 346\; \rm m \cdot s^{-1}$.

Let the frequency of this tune be $f\; \rm Hz$. The wavelength of the tune would be $\displaystyle \lambda = \frac{v}{f} \approx \frac{346}{f}$.

The distance between the first speaker and the listener is $12.608\; \rm m$. How many wavelengths can fit into that distance?

$\displaystyle \frac{12.608}{\lambda} \approx \frac{12.608}{346 / f} = \frac{12.608}{346} \cdot f$.

Similarly, the distance between the second speaker and the listener is $18.368\; \rm m$. Number of wavelengths in that distance:

$\displaystyle \frac{18.368}{\lambda} \approx \frac{18.368}{346 / f} = \frac{18.368}{346} \cdot f$.

Difference between these two numbers:

$\begin{aligned} &\frac{18.368}{346} \cdot f - \frac{12.608}{346} \cdot f = \frac{5.76}{346}\cdot f\end{aligned}$.

For destructive interference to occur, that difference should be equal to $\displaystyle \frac{1}{2}$, $\displaystyle 1 + \frac{1}{2}= \frac{3}{2}$, $\cdots$, or  $\displaystyle \left(k+ \frac{1}{2}\right)$ in general ($k$ can be any non-negative whole number.)

Let $\begin{aligned} \frac{5.76}{346}\cdot f = k + \frac{1}{2}\end{aligned}$, and solve for $f$.

$\begin{aligned} f&= \left.\left(k + \frac{1}{2}\right) \right/\frac{5.76}{346}\\ &= \frac{346}{5.76} \, k + \frac{1}{2} \times \frac{346}{5.76} \\ &\approx 60.1\, k + 30.0 \end{aligned}$.

The next step is to find the values of $k$ that ensure $20 \le f \le 20\times 10^{3}$ (frequency is between $20\; \rm Hz$ and $20\, \rm kHz$.) It turns out that $k = 0$ (the smallest $k$ value possible) would be sufficient. In that case, the frequency is approximately $30.0\; \rm Hz$. Using a larger $k$ would only increase the frequency.