Speed of the enemy submarine, $v_{ e }=360\, km / h =100 \,m / s$

Speed of sound in water, $v=1450 \,m / s$

The source is at rest and the observer (enemy submarine) is moving toward it. Hence, the apparen frequency ( $v^{\prime}$ ) received and reflected by the submarine is given by the relation:

$v^{\prime}=\left(\frac{v+v_{e}}{v}\right) \,v$

$=\left(\frac{1450+100}{1450}\right) \times 40=42.76\, kHz$

The frequency $\left(v^{\prime \prime}\right)$ received by the enemy submarine is given by the relation:

$v^{\prime \prime}=\left(\frac{v}{v+v_{s}}\right) v^{\prime}$

Where, $v_{ s }=100 \,m / s$

$\therefore v^{\prime \prime}=\left(\frac{1450}{1450-100}\right) \times 42.76=45.93 \,kHz$