In a calm sea, two submarines approach one another. Relative to the water, the first submarine moves at 36 km h ⁻¹ while the second submarine moves at 54 km h ⁻¹ . The first submarine emits a sound signal (sonar) with a frequency of 2000 Hz . Determine the frequency at which the second submarine receives this signal. Assume the speed of sound waves in water is 1500 m s ⁻¹ .

Step-by-step solution

Speed of the first submarine (source), v_s = 36 km h ⁻¹ = 36 5 18 m s ⁻¹ = 10 m s ⁻¹ Speed of the second submarine (observer), v_o = 54 km h ⁻¹ = 54 5 18 m s ⁻¹ = 15 m s ⁻¹ Speed of sound in water, v = 1500 m s ⁻¹ Actual frequency of the signal, f = 2000 Hz Using the Doppler effect formula for an approaching source and observer: f' = f ( v + v_o v - v_s ) Substituting the given values: f' = 2000 ( 1500 + 15 1500 - 10 ) f' = 2000 ( 1515 1490 ) f' = 303000 149 2033.56 Hz Rounding to the nearest integer gives 2034 Hz