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Mmohab · Answer 1 · 2021-11-22T08:27:03+0000

Answer:

Approximately
$30.0\; \rm Hz$ .

Step-by-step 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
$\displaystyle \lambda = (v)/(f) \approx (346)/(f)$ .

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

$\displaystyle (12.608)/(\lambda) \approx (12.608)/(346 / f) = (12.608)/(346) \cdot f$ .

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

$\displaystyle (18.368)/(\lambda) \approx (18.368)/(346 / f) = (18.368)/(346) \cdot f$ .

Difference between these two numbers:

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

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

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

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

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

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