Travelling along a straight line at a speed of 170 m s ⁻¹ , a source of sound emits a 1200 Hz note. A detector is located 200 m away from the source's line of motion. Determine the frequency of the sound received by the detector at the exact instant the source reaches its closest point to the detector. Assume the velocity of sound in air to be 340 m s ⁻¹ .

Step-by-step solution

Let the source move along the x-axis and the detector be at (0, d) . The source reaches the closest point to the detector at t = 0 . The sound received by the detector at this instant was emitted by the source at an earlier time -t₀ . At time -t₀ , the position of the source was (-v_s t₀, 0) . The distance between the source at the time of emission and the detector is D = (-v_s t₀)^2 + d^2 . The time taken for the sound to travel this distance is t₀ , so D = v t₀ . From the geometry, the angle between the velocity