Here, $\upsilon = 40.0\ \mathrm{kHz}$ ; Speed of sound wave in water, $v = 1, 450\ \mathrm{m\,s^{-1}}$ The frequency of the waves from SONAR system will undergo a change in the following two steps : (i) Frequency of waves from SONAR system as received by the enemysubmarine moving towards the system : Here, $v_{0} = 360\ \mathrm{km\,h^{-1}} = \frac{360 \times 1,000}{60 \times 60}$ $= 100\ \mathrm{m\,s^{-1}}$ Now, $\upsilon' = \frac{v+v_{0}}{v} \upsilon$ $= \frac{1,450+100}{1,450} \times 40.0 = 42.76\ \mathrm{kHz}$ (ii) Frequency of waves from the enemy’s submarine asreceived by SONAR system : The enemy submarine will reflect the waves of frequency $\upsilon'' = 42.756\ \mathrm{kHz}$ and will thus act as a source of waves moving with a speed; $v_s = 100\ \mathrm{m\,s^{-1}}$ toward the SONAR system (listener). If $\upsilon''$ is the apparent frequency as received by SONAR system, then $\upsilon'' = \frac{v}{v-v_{s}} \upsilon'$ $= \frac{1,450}{1,450-100} \times 42.76 = 45.93\ \mathrm{kHz}$