text: garnentas3m
  
  2022-01-03
  
  A bat locates insects by emitting ultrasonic “chirps” and then listening for echoes from the bugs. Suppose a bat chirp has a frequency of 25 kHz. How fast would the bat have to fly, and in what direction, for you to just barely be able to hear the chirp at 20 kHz?
  
  usumbiix
  
  Doppler effect:
  ${f}_{L}=\frac{v+{v}_{L}}{v+{v}_{S}}{f}_{S}$
  ${f}_{L}⇒$ Frequency observed by the listener, $v⇒$ Speed of sound,
  ${v}_{L}⇒$ Speed of listener, ${v}_{S}⇒$ Speed of the source of sound,
  ${f}_{S}⇒$ Frequency of the source of the sound.
  ${v}_{L}⇒$ is (((+))) when velocity of listener is from L(listener) to S(source),
  ${v}_{S}⇒$  is (((+))) when velocity of source is from L(listener) to (source).
  and the velocity is negative in the opposite situation.
  The frequency of the sound heard by the listener differs from the source frequency when the source and listener are moving apart from one another. When a car approaches you and sounds its horn, for instance, the pitch seems to get lower as the car passes.
  Apply: in most problems: we are asked to get the frequency of a certain sound observed by a moving observer from a moving source or a stationary one. The hard part is to determine the direction for each velocity and get the right signs.
  The bat must move away from me to make the frequency
  be heard at less frequency than the frequency from the bat
  ${v}_{L}=0$
  ${v}_{S}⇒$ is positive as the source is moving away from the listener
  (moving from L to S)
  ${f}_{L}=\frac{v+{v}_{L}}{v+{v}_{S}}{f}_{S}=\frac{v+0}{v+{v}_{bat}}{f}_{S}$,
  $v+{v}_{bat}=\frac{v{f}_{S}}{{f}_{L}}$
  ${v}_{bat}=\frac{v{f}_{S}}{{f}_{L}}-v=\frac{\left(343\frac{m}{s}\right)×\left(25000Hz\right)}{20000Hz}-343\frac{m}{s}=85.75\frac{m}{s}$
  $⇒$ Away from the listener
  
  Gerald Lopez
  
  a) $f‘={f}_{0}\left(\frac{u}{u+{v}_{s}}\right)$ Here u is velosity of sound $u=343\frac{m}{s}$
  $⇒20=25\left(\frac{343}{343+{v}_{s}}\right)$ ${f}_{0}=25kHz$ $f‘=20kHz$
  $⇒343×20+20{v}_{s}=343×25$
  $\therefore {v}_{s}=\frac{5×343}{20}=85.75\frac{m}{s}$
  Since appeared frequency is less than originil freq. direction of bat velosity will be toward you
  Note: There may be any other value of u. Please comment below in case of any correction.
  
  karton
  
  Sounds like this question is about the Doppler effect. If the source and receiver are in relative motion then the apparent frequency of the wave changes.
  When the motion decreases the separation between the source and the receiver, the frequency increases: the opposite is true when the distance increases. So in this case to decrease the inaudible 25 kHz chirp to the point where it can be heard requires motion of the source (the bat) away from the observer (answer 1)
  If v is the speed of sound in air ($340\frac{m}{s}$) and v_b is the speed of the bat then the frequency heard by the observer is ${f}_{obs}=\frac{vi}{v+{v}_{b}}{f}_{bat}$. Rearrange the Doppler formula to get ${v}_{b}=v\frac{\left({f}_{bat}-{f}_{obs}\right)}{{f}_{obs}}=85\frac{m}{s}$
  That's one fast bat ($85\frac{m}{s}=190$ m.p.h.)!
  
  Do you have a similar question?