Question

Hint Mary sings a note with a frequency of 888 Hz as she runs toward a wall at a speed of 4.17 m/s. What is the beat frequency between the note she sings and its Doppler-shifted reflection from the wall? In other words, what beat frequency does Mary detect between hearing the sound of her voice directly and hearing the sound reflected from the wall? Mary detects two frequencies. One is the frequency of her voice. The other is the Doppler-shifted frequency that occurs because of the reflection of the sound from the wall. When sound waves of two frequencies fi and /2 are detected simultaneously, the beat frequency is fVi-al (Equation 17.4.1). Two Doppler shifts must be accounted for. In the first, the wall is the observer and Mary is the moving source. Use fofs( + Vs/v) (Equation 16.4.2) to calculate the frequency observed" by the wall. In the second, Mary is the moving observer as she walks toward the wall, which is the source. Use fo-fs(1 + Vo/v) (Equation 16.4.1) to calculate the observed frequency I Mark for Review Confidence a A A A Submit Answer

Answer 1

Here, Marry will hear two frequencies. First her own frequency -

f1 = 888 Hz

The other frequency is reflected from the wall. Let us assume this frequency as f2 and calculate this.

Speed of Marry, v = 4.17 m/s

Speed of sound, c = 340 m/s

Apparant frequency received by the wall, f' = f1*(1 + v/c) = 888*(1 + 4.17/340) = 899 Hz

This frequency acts as a source for the reflected wave.

So, frequency of the reflected sound wave received by Marry -

f2 = f'*(1 + v/c) = 899*(1+4.17/340) = 910 Hz

Therefore, the beat frequency = f2 - f1 = 910 - 888 = 22 Hz (Answer)

Hope, you understand the solution!