Answer
(a) $35.3KHz$
(b) higher
(c) $35.4KHz$
Work Step by Step
(a) We can find the frequency heard by the moth as follows:
$f^{\prime}=f(\frac{v-v_{\circ}}{v-v_s})$
We plug in the known values to obtain:
$f^{\prime}=(35\times 10^3Hz)(\frac{373m/s-0}{373m/s-3.25m/s})$
$f^{\prime}=35334.6Hz$
$f^{\prime}=35.3KHz$
(b) We know that the frequency heard by the moth will be larger if the speed of the bat is increased because in this case the source is approaching the stationary observer.
(c) We know that
$f^{\prime}=f(\frac{v-v_{\circ}}{v-v_s})$
We plug in the known values to obtain:
$f^{\prime}=(35\times 10^3Hz)(\frac{343m/s-0}{343m/s-4.25m/s})$
$f^{\prime}=35.4KHz$
