<p>A bat flies toward a wall, emitting a steady sound with a frequency of 20.0 k

Question

<p>A bat flies toward a wall, emitting a steady sound with a frequency of 20.0 kHz. This bat hears its own sound plus the sound reflected by the wall, how fast should the bat fly, Vb, to hear a beat frequency of 180 Hz?<br /><br />Take the speed of sound to be 344 m/s.</p>
<p>Vb= ? m/s</p>

Explanation / Answer

Beat frequency = f2 - f0 225 = f2 - 20000 f2 = 20225 Hz The difference in frequency is due to the Doppler effect. When the sound hits the wall, wall acts as the still observer and the bat acts as the source moving towards the observer. To this movement the appropriate Doppler effect equation is. f1 = V / ( V - Vs ) x fo f1 - frequency hits the wall V - velocity of sound Vs - velocity of the source ( bat ) fo - frequency of the source f1 = 344/(344-Vs) x 20000 --> [1] When the bat hears the reflected sound, bat acts as the observer moving towards the source and the wall acts as the still source. To this movement the appropriate Doppler effect equation is. f2 = (V + Vo) / V x f1 f2 - frequency heard by the bat V - velocity of sound Vo - velocity of the observer ( bat ) fo - frequency reflected from the wall 20225 = (344+Vo)/344 x f1 --> [2] plug in f1 from [1] 20225 = (344+Vo)/344 x [344/(344-Vs) x 20000] Vo = Vs so, 20225 = (344+Vo)/344 x [344/(344-Vo) x 20000] 20225 / 20000 = (344+Vo) / (344-Vo) solve for Vo Vo = 1.92418 ms^-1

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