Answer
$0.68KHz; 2.0KHz$
Work Step by Step
We can find the required frequencies as follows:
$\Delta d=\frac{\lambda_1}{2}$
$\implies 0.25m=\frac{\lambda_1}{2}$
$\lambda_1=0.50m$
Now $f_1=\frac{v}{\lambda_1}$
$\implies f_1=\frac{343m/s}{0.502m}=6.8\times 10^2Hz=0.68KHz$
Similarly $\lambda_2=\frac{2}{3}(0.25m)=0.167m$
and $f_2=\frac{v}{\lambda_2}$
We plug in the known values to obtain:
$f_2=\frac{343m/s}{0.167m}$
$f_2=2.0\times 10^3Hz=2KHz$
