Respuesta :
Answer:
Speed at which the bat have to fly is 90.895 m/s away from the human (listener) in positive direction.
Explanation:
Given:
Frequency of the bat that is source here, [tex]f_S[/tex] = 25.3 kHz = 25.3 * 10^3 Hz
Frequency of the listener (human), [tex]f_L[/tex] = 20 kHz = 20*10^3 Hz
We have to identify how fast the bat have to fly in order for a person to hear these chirps .
Let the velocity of bat that is source is "Vs" and "Vs" = "Vbat".
Doppler effects formulae :
- When the source is receding (moving away) [tex]f_L=(\frac{V+V_L}{V+V_S}) f_S[/tex]
- When the source is approaching [tex]f_L=(\frac{V+V_L}{V-V_S}) f_S[/tex]
- Speed of sound in [tex]V_S =433.895-343[/tex]the air (medium), [tex]V = 343\ ms^-^1[/tex]
Using the above formula and considering that the bat is moving away so that the human can listen the chirps also [tex]V_L=0[/tex] as listener is stationary.
⇒ [tex]f_L=(\frac{V+V_L}{V+V_S}) f_S[/tex] ⇒ [tex]f_L=(\frac{V+0}{V+V_S}) f_S[/tex]
Re-arranging in terms of Vs.
⇒ [tex]V+V_S =\frac{V\times f_S}{f_L}[/tex]
⇒ [tex]V_S =\frac{V\times f_S}{f_L}-V[/tex]
⇒ [tex]V_S =\frac{343\times 25.3\times 10^3}{20\times 10^3}-343[/tex]
⇒ [tex]V_S=90.895[/tex] m/s
The speed at which the bat have to fly is 90.895 m/s away from the human (listener) in positive direction.
