Answer
(a) $0.344 m$
(b) $1.2 \times 10^{-5} m$
(c) 50 Hz
Work Step by Step
(a) From $\nu = f\lambda$, we get $\lambda = \frac{\nu}{f} = \frac{344}{1000} = 0.344 m$.
(b) $p_{max} = BkA$. For constant $Bk$, we get $\frac{p_{max1}}{A_1} = \frac{p_{max2}}{A_2}$
$ \implies A_2 = A_1(\frac{p_{max2}}{p_{max1}}) = (1.2\times 10^{-8})(\frac{30}{3\times 10^{-2}}) = 1.2 \times 10^{-5} m$
(c) $p_{max} = BkA = \frac{2\pi BkA}{\lambda}$. Again, $2\pi BkA$ is constant, so we have $p_{max1} \lambda_{1} = p_{max2} \lambda_{2}$.
$\implies \lambda_{2} = \lambda_1(\frac{p_{max1}}{p_{max2}}) = (0.344)(\frac{3\times 10^{-2}}{1.5\times 10^{-3}}) = 6.9 m$.
Then $f = \nu / \lambda = \frac{344}{6.9} = 50 Hz$.
