Answer
(a) $3.6KHz;9.6cm$
(b) $11KHz;3.2cm$
(c) greater than
Work Step by Step
(a) We can find the required fundamental frequency and wavelength as follows:
$f_1=\frac{v}{4L}$
We plug in the known values to obtain:
$f_1=\frac{343m/s}{4\times 2.4\times 10^{-2}m}$
$f_1=3.57\times 10^3Hz=3.6KHz$
and $\lambda_1=4L$
$\lambda_1=4(2.4cm)=9.6cm$
(b) We can find the required frequency and wavelength as follows:
$f_3=3f_1$
$f_3=3(3.6KHz)=11KHz$
and $\lambda_3=\frac{\lambda_1}{3}$
$\lambda_3=\frac{9.6cm}{3}=3.2cm$
(c) We know that a shorter wavelength has a greater frequency; thus, if the person's ear canal is shorter than $L=2.4cm$, then the fundamental frequency will be greater.