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Sagot :
The bat should fly about 0.8065 m/s in order to hear a beat frequency of 8 Hz.
The shift in a wave's frequency caused by an observer moving away from the wave source is known as the Doppler effect or Doppler shift. The equation for this is given by,
[tex]f=\frac{v \pm v_0}{v\mp v_s}f_s[/tex]
where f is the observed frequency, v is the speed or velocity of the sound waves, v₀ is observer velocity ("+" if it's towards and "-" if it's away from the source), vs is source velocity("-" if it's towards and "+" if it's away from the observer), and fs is the actual frequency of the sound waves.
The velocity of sound waves in the air is 344 m/s, v₀ is zero for the incident, v is the source velocity towards the wall, and fs is 1700 Hz.
Then, the incident frequency is given by,
[tex]f_{incidence}=\frac{344-0}{344-v}\times f_s[/tex]
The wall is stationary, therefore, f(incidence)=f(reflected). Then, the new frequency the bat hears is,
[tex]\begin{aligned}f_{new}&=\frac{344+v}{344}\times f_{reflected}\\&=\frac{344+v}{344}\times \frac{344-0}{344-v}\times f_s\\&=\frac{344+v}{344-v} \times f_s\end{aligned}[/tex]
To hear a beat frequency of 8 Hz, the velocity of the bat should be,
[tex]\begin{aligned}f_{new}-f_s&=8\\\left(\frac{344+v}{344-v} \right)f_s-f_s&=8\\\frac{344+v}{344-v}-1&=\frac{8}{f_s}\\\frac{344+v-344+v}{344-v}&=\frac{8}{1700}\\2v&=0.0047(344-v)\\2v&=1.6168-0.0047v\\2.0047v&=1.6168\\v&=\mathrm{0.8065\;m/s}\end{aligned}[/tex]
The answer is 0.8065 m/s.
To know more about the Doppler effect:
https://brainly.com/question/15318474
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