In a calm sea, two submarines approach one another. Relative to the...

Speed of the first submarine, $v_1 = 36 \text{ km h}^{-1} = 10 \text{ m s}^{-1}$

Speed of the second submarine, $v_2 = 54 \text{ km h}^{-1} = 15 \text{ m s}^{-1}$

Speed of sound in water, $v = 1500 \text{ m s}^{-1}$

Frequency of the emitted signal, $f_0 = 2000 \text{ Hz}$

The frequency received by the second submarine is given by the Doppler effect formula for an observer moving towards a source and a source moving towards the observer:

$f_1 = f_0 \left( \dfrac{v + v_2}{v - v_1} \right)<br><br>The second submarine reflects this signal, acting as a moving source, while the first submarine acts as a moving observer. The frequency of the reflected signal received by the first submarine is:<br><br>$f_2 = f_1 \left( \dfrac{v + v_1}{v - v_2} \right)$<br><br>Substituting$f_1$into the equation for$f_2$:<br><br>$f_2 = f_0 \left( \dfrac{v + v_2}{v - v_1} \right)\left( \dfrac{v + v_1}{v - v_2} \right)

Substituting the given values:

$f_2 = 2000 \left( \dfrac{1500 + 15}{1500 - 10} \right) \left( \dfrac{1500 + 10}{1500 - 15} \right)$

$f_2 = 2000 \left( \dfrac{1515}{1490} \right) \left( \dfrac{1510}{1485} \right)<br><br>$f_2 = 2000 \left( \dfrac{303}{298} \right) \left( \dfrac{302}{297} \right)

f_2 = 2000 \left(\dfrac{101}{99} \right)\left( \dfrac{151}{149} \right)$<br><br>$f_2 = 2000 \left( \dfrac{15251}{14751} \right)

$f_2$\approx$2067.79$\text{ Hz}

Rounding to the nearest integer, the frequency is $2068 \text{ Hz}.$