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

Speed of the first submarine (source), $v_s = 36 \text{ km h}^{-1} = 36 \times \dfrac{5}{18} \text{ m s}^{-1} = 10 \text{ m s}^{-1}$

Speed of the second submarine (observer), $v_o = 54 \text{ km h}^{-1} = 54 \times \dfrac{5}{18} \text{ m s}^{-1} = 15 \text{ m s}^{-1}$

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

Actual frequency of the signal, $f = 2000 \text{ Hz}$

Using the Doppler effect formula for an approaching source and observer:

f' = f \left(\dfrac{v + v_o}{v - v_s} \right)

Substituting the given values:

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

f' = 2000 \left(\dfrac{1515}{1490} \right)

$f' =$\dfrac{303000}{149} \approx$2033.56 \text{ Hz}$

Rounding to the nearest integer gives $2034 \text{ Hz}.$