Frequency of the SONAR system is given as $\mathrm{f}=40 \mathrm{kHz}=40 \times 10^{3} \mathrm{~Hz}$

Speed of sound in water is given as $v=1450 \mathrm{~m} / \mathrm{s}$

Speed of the enemy submarine is given as

$\mathrm{v}_{0}=360 \mathrm{~km} / \mathrm{h}=360 \times(5 / 18)=100 \mathrm{~m} / \mathrm{s}$

The energy submarine approaches the SONAR, which is at rest. As a result, the perceived frequency is determined by the relationship.

$\mathrm{f}^{\prime}=\left[\left(\mathrm{v}+\mathrm{v}_{0}\right) / \mathrm{v}\right] \mathrm{f}$

$\begin{array}{l}
=[(1450+100) / 1450] \times 40 \times 10^{3} \\
=42.75 \times 10^{3}
\end{array}$

This frequency $\left(f^{\prime}\right)$ is reflected by the enemy submarine and it is observed by the SONAR. So, $v_{s}=360 \mathrm{km} / \mathrm{s}=100 \mathrm{~m} / \mathrm{s}$

$\begin{array}{l}
\\
f^{\prime \prime}=\left(\mathrm{v} /\left(\mathrm{v}-\mathrm{v}_{\mathrm{s}}\right)\right) \times \mathrm{f}^{\prime} \\
=(1450 /(1450-100)) \times 42.75 \times 10^{3} \\
=(1450 / 1350) \times 42.75 \times 10^{3} \\
=45.91 \times 10^{3}
\end{array}$