Two stereo speakers are separated by a distance of 2.40 m . A...

Let the two speakers be $S_1$ and $S_2$, and the listener be at point $P$.

The distance from the first speaker to the listener is:
$L_1 = 3.20 \text{ m}$

The distance from the second speaker to the listener is:
$L_2 = \sqrt{(2.40)^2 + (3.20)^2} = \sqrt{5.76 + 10.24} = \sqrt{16.00} = 4.00 \text{ m}$

The path difference $\Delta x$ between the sound waves reaching the listener is:
$\Delta x = L_2 - L_1 = 4.00 - 3.20 = 0.80 \text{ m}$

For minimum sound intensity (destructive interference), the path difference must be an odd multiple of half the wavelength:
$\Delta x = (2n + 1)\dfrac{\lambda}{2}$, where $n = 0, 1, 2, \dots$

Using the relation $v = f\lambda$, we have $\lambda = \dfrac{v}{f}$. Substituting this into the condition for destructive interference:
$\Delta x = (2n + 1)\dfrac{v}{2f}$

Solving for the frequency $f$:
$f = (2n + 1)\dfrac{v}{2\Delta x}$

Given the speed of sound $v = 320 \text{ m s}^{-1}$ and $\Delta x = 0.80 \text{ m}$:
$f = (2n + 1)\dfrac{320}{2 \times 0.80} = (2n + 1)\dfrac{320}{1.60} = 200(2n + 1) \text{ Hz}$

The listener can hear frequencies in the audible range $20 \text{ Hz} \le f \le 2000 \text{ Hz}$:
$20 \le 200(2n + 1) \le 2000$

Dividing by $200$:
$0.1 \le 2n + 1 \le 10$

Subtracting $1$:
$-0.9 \le 2n \le 9$

Dividing by $2$:
$-0.45 \le n \le 4.5$

Since $n$ must be a non-negative integer, the possible values are $n = 0, 1, 2, 3, 4$.

Thus, the frequencies for minimum sound intensity are $200(2n + 1) \text{ Hz}$, where $n = 0, 1, 2, 3, 4$.