Answer
$3.86 \times 10^{3} Hz$
Work Step by Step
We know that $p_{max} = BkA$. Since $k = 2\pi /\lambda = 2\pi f /\nu$, $p_{max} = \frac{2\pi fBA}{\nu}$.
For $\nu = 344 m/s$, we have . . .
$f = \frac{\nu p_{max}}{2\pi BA} = \frac{(344)(10)}{2\pi (1.42 \times 10^{5})(10^{-6})} = 3.86 \times 10^3 Hz$.
