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# 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.

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### James

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## A bat locates insects by emitting ultrasonic “chirps” and then listeni

Solved: 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

## A bat locates insects by emitting ultrasonic “chirps” and then listeni

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?

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Author has 33 answers

Doppler effect fL=v+vLv+vSfS fL⇒

Frequency observed by the listener,

v⇒ Speed of sound, vL⇒ Speed of listener, vS⇒

Speed of the source of sound,

fS⇒

Frequency of the source of the sound.

vL⇒

is (((+))) when velocity of listener is from L(listener) to S(source),

vS⇒

is (((+))) when velocity of source is from L(listener) to (source).

and the velocity is negative in the opposite situation.

When a source of sound and a listener are in motion relative to each other, the frequency of the sound heard by the listener is not the same as the source frequency. For example, when a car approaches you with its horn sounding, the pitch seems to drop 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

vL=0 vS⇒

is positive as the source is moving away from the listener

(moving from L to S)

fL=v+vLv+vSfS=v+0v+vbatfS

, v+vbat=vfSfL

vbat=vfSfL−v=(343ms)×(25000Hz)20000Hz−343ms=85.75ms

Away from the listner(me)

### Not exactly what you’re looking for?

Ask My Question This is helpful 0 Gerald Lopez Answered 2022-01-05

Author has 29 answers

a) f‘=f0(uu+vs)

Here u is velosity of sound

u=343ms ⇒20=25(343343+vs) f0=25kHz f‘=20kHz ⇒343×20+20vs=343×25 ∴vs=5×34320=85.75ms

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.

This is helpful 0 karton Answered 2022-01-10

Author has 368 answers

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 (

340ms

) and v_b is the speed of the bat then the frequency heard by the observer is

fobs=viv+vbfbat

. Rearrange the Doppler formula to get

vb=v(fbat−fobs)fobs=85ms

That's one fast bat (

85ms=190 m.p.h.)! This is helpful 0

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Source : plainmath.net

## A bat locates insects by emitting ultrasonic “chirps” and th

Find step-by-step Physics solutions and your answer to the following textbook question: 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?.

### Question

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?

### Explanation

Verified

\begin{gathered} {\text{Doppler effect:}} \\ \\ {f_L} = \frac{{v + {v_L}}}{{v + {v_S}}}{f_S} \\ \\ {f_L} \Rightarrow {\text{Frequency observed by the listener}},v \Rightarrow {\text{Speed of sound,}} \\ {v_L} \Rightarrow {\text{Speed of listener}},{v_S} \Rightarrow {\text{Speed of the source of sound,}} \\ {f_S} \Rightarrow {\text{Frequency of the source of the sound}}{\text{.}} \\ \\ {v_L} \to {\text{ is ((( + ))) when velocity of listener is from L(listener) to S(source),}} \\ {v_S} \to {\text{ is (((}} + ))){\text{ when velocity of source is from L(listener) to S(source)}}{\text{.}} \\ {\text{and the velocity is negative in the opposite situation}}{\text{.}} \\ \end{gathered}

Doppler effect: f L ​ = v+v S ​ v+v L ​ ​ 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 S(source).

and the velocity is negative in the opposite situation.

When a source of sound and a listener are in motion relative to each other, the frequency of the sound heard by the listener is not the same as the source frequency. For example, when a car approaches you with its horn sounding, the pitch seems to drop 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.

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### Related questions

QUESTION

Although bats are not known for their eyesight, they are able to locate prey (mainly insects) by emitting high-pitched sounds and listening for echoes. A paper appearing in Animal Behaviour (“The Echolocation of Flying Insects by Bats” [1960]: 141– 154) gave the following distances (in centimeters) at which a bat first detected a nearby insect: a. Compute the sample mean distance at which the bat first detects an insect. b. Compute the sample variance and standard deviation for this data set. Interpret

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Dolphins and bats determine the location of their prey by emitting ultrasonic sound waves. Which wave phenomenon is involved?

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Source : quizlet.com

## SOLVED: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 \mathrm{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 \mathrm{kHz} ?

VIDEO ANSWER: um, if you think about bats and this is in, ah, in a time period where you know we have virus is going on Onda And you know, one of the viruses that everyone's talking about is the coffee 19 virus. Fo