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 , which is then reflected by the second submarine. Determine the frequency of the reflected signal as received by the first submarine. Assume the speed of sound waves in water is

Step-by-step solution

Speed of the first submarine, v₁ = 36 km h ⁻¹ = 10 m s ⁻¹ Speed of the second submarine, v₂ = 54 km h ⁻¹ = 15 m s ⁻¹ Speed of sound in water, v = 1500 m s ⁻¹ Frequency of the emitted signal, f₀ = 2000 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: f₁ = f₀ ( v + v₂ v - v₁ ) The second submarine reflects this signal, acting as a moving source, while the first submarine acts as a moving observer.