(a) Here, $v_{s} = 0\ and\ v_{L}$ is negative where $v_{s}$is velocity of source and $v_{L}$is velocity of listener (aeroplane)
[image:15_image60.emf]
If apparent frequency is v’ and v is actual frequency, then
$v^{''} = \frac{v - \left( - v_{L} \right)}{v}v = \frac{v + v_{L}}{v}L$
$i.e.,$ v’>v
So, apparent frequency will increase, it means apparent wavelength will decrease.