Frequency of SONAR, $v=40 \mathrm{kHz}$ $=40 \times 10^3 \mathrm{~Hz}$
Speed of observer $=v_o=360 \mathrm{~km} / \mathrm{h}=100 \mathrm{~m} / \mathrm{s}$
Speed of sound $=v=1450 \mathrm{~m} / \mathrm{s}$.
$\because \quad$ The source is at rest and observer is moving towards the source
$\therefore \quad$ Apparent frequency received by the submarine $=$ $v^{\prime}=\frac{\left(v+v_o\right) v}{v}$ $=\frac{(1450+100) \times 40 \times 10^3}{1450}=4.276 \times 10^4 \mathrm{~Hz}$
This frequency is reflected by the enemy submarine and is observed by SONAR.
$\therefore \quad v_s=360 \mathrm{~km} / \mathrm{s}=100 \mathrm{~m} / \mathrm{s}, v_o=0$
$\therefore \quad$ Apparent frequency $=v^{\prime \prime}=\frac{v \times v}{v-v_s}$
$$
=\frac{1450 \times 4.276 \times 10^4}{1450-10}=4.59 \times 10^4 \mathrm{~Hz} .
$$