A bird is flying directly toward a stationary bird-watcher and emits a frequency of 1390 Hz. The bird-watcher, however, hears a frequency of 1430 Hz. What is the speed of the bird, expressed as a percentage of the speed of sound?
Physics - Elena, Sunday, March 3, 2013 at 3:18pm
v/u=(f-f₀)/f= (1430-1390) /1430=2.8•10⁻² m/s
v/u = 2.8%