The Doppler Effect - frequency of a bat reflected on wall

In summary, the conversation discusses the Doppler Effect in relation to a bat flying towards a wall and emitting a sound with a frequency of 22.0 kHz. The question asks how fast the bat should fly to hear a beat frequency of 190 Hz. Suggestions are made to approach the problem using wavelength and the formula f_{L}=[(v+v_{L})/(v+v_{s})]f_{s}, but the equations and variables involved are causing confusion.
  • #1
forestmine
203
0
The Doppler Effect -- frequency of a bat reflected on wall

Homework Statement



A bat flies toward a wall, emitting a steady sound with a frequency of 22.0 kHz. This bat hears its own sound plus the sound reflected by the wall.

How fast should the bat fly to hear a beat frequency of 190 Hz?

Homework Equations



Doppler Effect equations

The Attempt at a Solution



I know this is a case of the Doppler effect, but I don't understand the variations of the equations in terms of when the source is stationary, listener is moving towards it, etc. I also understand that the frequency the bat emits will be different than the frequency rebounded by the wall. I kind of need to be walked through this one, because all of the explanations I've seen thus far aren't really getting through...

Thanks!
 
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  • #2


forestmine said:

Homework Statement



A bat flies toward a wall, emitting a steady sound with a frequency of 22.0 kHz. This bat hears its own sound plus the sound reflected by the wall.

How fast should the bat fly to hear a beat frequency of 190 Hz?

Homework Equations



Doppler Effect equations

The Attempt at a Solution



I know this is a case of the Doppler effect, but I don't understand the variations of the equations in terms of when the source is stationary, listener is moving towards it, etc. I also understand that the frequency the bat emits will be different than the frequency rebounded by the wall. I kind of need to be walked through this one, because all of the explanations I've seen thus far aren't really getting through...

Thanks!

you could try a first principles approach via wavelength.

The reflected wave is to have a frequency 190 higher [it will be higher not lower as the bat is moving towards the wall]
So the two frequencies detected are 22000 Hz and 22190 Hz

calculate the wavelength of each of those waves.

Lets assume they are 15 mm and 14 mm

In order to achieve that apparent reduction, the bat would have to move forward 1mm between each cycle of sound produced.
That means 1 mm every 22000th of a second. Scale that up and you will have a speed of 22 m/s. I doubt very much that the real wavelengths differ that much, so the real velocity will be less than that.
 
  • #3


So can I just use the speed of sound in the air for my velocity in order to find the wavelength?
 
  • #4


The solution shows use of the equation f[itex]_{L}[/itex]=[(v+v[itex]_{L}[/itex])/(v+v[itex]_{s}[/itex])]f[itex]_{s}[/itex]

Could you perhaps explain it in terms of this equation? I understand that the positive direction will be from the listener to the source, so in the direction that the bat is flying. But I'm not entirely sure what I ought to be using for each variable.

The solution doesn't show any use of wavelengths at all, so I'm a bit confused.
 
  • #5


forestmine said:
The solution shows use of the equation f[itex]_{L}[/itex]=[(v+v[itex]_{L}[/itex])/(v+v[itex]_{s}[/itex])]f[itex]_{s}[/itex]

Could you perhaps explain it in terms of this equation? I understand that the positive direction will be from the listener to the source, so in the direction that the bat is flying. But I'm not entirely sure what I ought to be using for each variable.

The solution doesn't show any use of wavelengths at all, so I'm a bit confused.

Certainly that formula uses the speed of sound and the speed of "L" and "s" , where L is the listener and s is the source.

The bat emits a sound - it is the source, the wall received the sound - it is a pseudo listener.

The pitch of sound will arrive as a higher pitch.

That sound is then reflected: the wall becomes the source and the bat the listener.

The bat will detect that reflected sound [already a higher pitch that 22000] as an even higher pitch.

Looks like you analyse it twice - and must have been given a speed of sound to use??



My suggestion of using wavelengths was, as stated, to use first principles. An analysis of a general example will lead to a velocity expression - like this one shown here.
 
  • #6


Meh, ignore this post, I misread the initial problem. Sorry!
 
  • #7


It seems I'm still doing something wrong. If I observe the instance in which the wall is the listener, I have f[itex]_{w}[/itex]=(v+v[itex]_{w}[/itex])/v-v[itex]_{b}[/itex])f[itex]_{b}[/itex]. Plugging in, 11700=(344/344-v[itex]_{b}[/itex])(1700). Which gives me a velocity of 294.02m/s, given that v[itex]_{b}[/itex] is my only unknown. But that means I've completely neglected the situation in which the bat is the listener and the wall is the source. I don't really see how to combine the two.

For that matter, I'm confused by the equations I'm supposed to use and what variables they entail.

For f heard by the wall I know it's [(v+v[itex]_{w}[/itex])/(v-v[itex]_{b}[/itex])]f[itex]_{b}[/itex], as mentioned above. And for the frequency heard by the bat, it's [(v+v[itex]_{b}[/itex])/(v-v[itex]_{w}[/itex])]f[itex]_{w}[/itex], where f[itex]_{w}[/itex] is the reflected, higher frequency.

I'm so confused!
 
  • #8


forestmine said:
It seems I'm still doing something wrong. If I observe the instance in which the wall is the listener, I have f[itex]_{w}[/itex]=(v+v[itex]_{w}[/itex])/v-v[itex]_{b}[/itex])f[itex]_{b}[/itex]. Plugging in, 11700=(344/344-v[itex]_{b}[/itex])(1700). Which gives me a velocity of 294.02m/s, given that v[itex]_{b}[/itex] is my only unknown. But that means I've completely neglected the situation in which the bat is the listener and the wall is the source. I don't really see how to combine the two.

For that matter, I'm confused by the equations I'm supposed to use and what variables they entail.

For f heard by the wall I know it's [(v+v[itex]_{w}[/itex])/(v-v[itex]_{b}[/itex])]f[itex]_{b}[/itex], as mentioned above. And for the frequency heard by the bat, it's [(v+v[itex]_{b}[/itex])/(v-v[itex]_{w}[/itex])]f[itex]_{w}[/itex], where f[itex]_{w}[/itex] is the reflected, higher frequency.

I'm so confused!

Where did that frequency value of 11700 come from?
 
  • #9


Ah, my mistake! I'm working a problem with slightly different given frequencies (than the ones I've posted) so that I'd ensure I would figure out how to work it in another instance.

That being said, ignore 11700. So, for this problem, it would be 22190=(the velocity part written above)x22000?
 
  • #10


forestmine said:
It seems I'm still doing something wrong. If I observe the instance in which the wall is the listener, I have f[itex]_{w}[/itex]=(v+v[itex]_{w}[/itex])/v-v[itex]_{b}[/itex])f[itex]_{b}[/itex]. Plugging in, 11700=(344/344-v[itex]_{b}[/itex])(1700). Which gives me a velocity of 294.02m/s, given that v[itex]_{b}[/itex] is my only unknown. But that means I've completely neglected the situation in which the bat is the listener and the wall is the source. I don't really see how to combine the two.

For that matter, I'm confused by the equations I'm supposed to use and what variables they entail.

For f heard by the wall I know it's [(v+v[itex]_{w}[/itex])/(v-v[itex]_{b}[/itex])]f[itex]_{b}[/itex], as mentioned above. And for the frequency heard by the bat, it's [(v+v[itex]_{b}[/itex])/(v-v[itex]_{w}[/itex])]f[itex]_{w}[/itex], where f[itex]_{w}[/itex] is the reflected, higher frequency.

I'm so confused!

Also, I suspect this can be done in one go.

When you approach a mirror at 3 m/s, you image approaches you at 6 m/s [that is all with light]

Acoustically I think this is similar.

We have a bat approaching a wall at speed V. That means its image is approaching it at speed 2V.

The bat, and the bats image are producing a frequency of 22000 Hz.
The bat hears its own frequency, plus the frequency form the "approaching bat" with the doppler shift resulting in 190 beats per second.
 
  • #11


Ok, that much makes sense, although, I still do not at all see how to go about solving this.
 

Related to The Doppler Effect - frequency of a bat reflected on wall

1. What is the Doppler Effect?

The Doppler Effect is a phenomenon that occurs when there is a change in the frequency of a wave due to the relative motion between the source of the wave and the observer.

2. How does the Doppler Effect apply to bats?

Bats use echolocation to navigate and hunt for prey. They emit high frequency sound waves that bounce off objects and return to the bat, allowing them to create a mental map of their surroundings. The Doppler Effect helps bats determine the direction and speed of their prey by analyzing the changes in frequency of the reflected sound waves.

3. Why is the frequency of a bat's sound waves important?

The frequency of a bat's sound waves is important because it determines the distance and speed of the bat's prey. Higher frequency sound waves have shorter wavelengths and are better for detecting small objects, while lower frequency sound waves have longer wavelengths and are better for detecting larger objects.

4. How does the Doppler Effect affect the frequency of a bat's sound waves?

When a bat is flying towards its prey, the sound waves it emits have a higher frequency due to the compression of the waves. As the bat flies away from its prey, the sound waves have a lower frequency due to the stretching of the waves. This change in frequency is known as the Doppler Effect.

5. How is the Doppler Effect used to study bat behavior?

By analyzing the frequency of a bat's sound waves, scientists can determine the behavior and movements of bats. For example, a higher frequency may indicate that a bat is flying towards its prey, while a lower frequency may indicate that it is flying away. This information can help researchers understand how bats navigate and hunt in different environments.

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