A bat flies at a speed of 6 m/s between two parallel walls while emitting an ultrasonic wave of frequency 4.5 10^4 Hz . Determine the two frequencies heard by the bat, as well as the beat frequency between them. The speed of sound is 330 m/s .

Step-by-step solution

The bat acts as both a moving source and a moving observer. For the wall the bat is approaching, the frequency of the reflected wave heard by the bat is given by the Doppler effect formula: f₁ = f ( v + v_b v - v_b ) Substituting v = 330 m/s , v_b = 6 m/s , and f = 4.5 10^4 Hz : f₁ = 4.5 10^4 ( 330 + 6 330 - 6 ) = 4.5 10^4 ( 336 324 ) = 4.5 10^4 ( 28 27 ) 4.67 10^4 Hz For the wall the bat is moving away from, the frequency of the reflected wave heard by the bat is: f₂ = f ( v - v_b v + v_b ) f₂ = 4.5 10^4 ( 330 - 6