A bat and an insect Doppler's Effect problem.

[kudoushinichi88]

The problems in this set of problems that I am doing are generally hard (at least to me). Here's the 5th question which I'm not sure whether or not it IS that simple or just deceptively simple.

Horseshoe bats emit sounds from their nostrils and then listen to the frequency of the sound reflected from their prey to determine the prey's speed. If a horseshoe bat flying at speed v_bat emits sound of frequency f_bat, the sound it hears reflected from an insect flying towards the bat has a higher frequency f_refl. What is the speed of the insect?

Since f_refl has a higher frequency, the insect is moving towards the bat. Therefore,

[tex]f_{refl}=\frac{v+v_{bat}}{v-v_{insect}}f_{bat}[/tex]

which simply leads to

[tex]v_{insect}=v-(v+v_{bat})\frac{f_{bat}}{f_{refl}}[/tex]

in which v here is the speed of sound.
Am I missing something or I'm just paranoid after studying non-stop for my finals since the past week?

 

[alphysicist]

Hi kudoushinichi88,

The problems in this set of problems that I am doing are generally hard (at least to me). Here's the 5th question which I'm not sure whether or not it IS that simple or just deceptively simple.

Horseshoe bats emit sounds from their nostrils and then listen to the frequency of the sound reflected from their prey to determine the prey's speed. If a horseshoe bat flying at speed v_bat emits sound of frequency f_bat, the sound it hears reflected from an insect flying towards the bat has a higher frequency f_refl. What is the speed of the insect?

Since f_refl has a higher frequency, the insect is moving towards the bat. Therefore,

[tex]f_{refl}=\frac{v+v_{bat}}{v-v_{insect}}f_{bat}[/tex]

The right hand side of this expression is not the reflected frequency that the bat hears; it is the frequency that the insect "hears" (that is, it is the frequency that the insect detects).

So now, using that, you have to determine what is the reflected frequency that the bat hears.

which simply leads to

[tex]v_{insect}=v-(v+v_{bat})\frac{f_{bat}}{f_{refl}}[/tex]

in which v here is the speed of sound.



Am I missing something or I'm just paranoid after studying non-stop for my finals since the past week?

 

[kudoushinichi88]

Okay, so the frequency of the sound reflected off the insect isn't the sound heard by the bat. Therefore, I applied the Doppler's Effect equation twice.

[tex]
f_{ins}=\frac{v+v_{ins}}{v-v_{bat}}f_{bat}
[/tex]

Where [itex]f_{ins}[/itex] and [itex]v_{ins}[/itex] are frequency heard by insect and speed of insect respectively.

Therefore, frequency heard by bat [tex]f_{refl}[/itex] is,

[tex]
f_{refl}=\frac{v+v_{bat}}{v-v_{ins}}f_{ins}
[/tex]

which leads to

[tex]
f_{refl}=\frac{v+v_{bat}}{v-v_{ins}}\cdot\frac{v+v_{ins}}{v-v_{bat}}f_{bat}
[/tex]

Rearranging, I get a horrible equation of

[tex]
v_{ins}=\frac{f_{refl}v(v-v_{bat})-f_{bat}v(v+v_{bat})}{f_{refl}(v-v_{bat})-f_{bat}(v+v_{bat})}[/tex]

does this make more sense?

 

[alphysicist]

Okay, so the frequency of the sound reflected off the insect isn't the sound heard by the bat. Therefore, I applied the Doppler's Effect equation twice.

[tex]
f_{ins}=\frac{v+v_{ins}}{v-v_{bat}}f_{bat}
[/tex]

Where [itex]f_{ins}[/itex] and [itex]v_{ins}[/itex] are frequency heard by insect and speed of insect respectively.

Therefore, frequency heard by bat [tex]f_{refl}[/itex] is,

[tex]
f_{refl}=\frac{v+v_{bat}}{v-v_{ins}}f_{ins}
[/tex]

which leads to

[tex]
f_{refl}=\frac{v+v_{bat}}{v-v_{ins}}\cdot\frac{v+v_{ins}}{v-v_{bat}}f_{bat}
[/tex]

This looks right to me.

Rearranging, I get a horrible equation of

[tex]
v_{ins}=\frac{f_{refl}v(v-v_{bat})-f_{bat}v(v+v_{bat})}{f_{refl}(v-v_{bat})-f_{bat}(v+v_{bat})}[/tex]

I think you have a sign error here. An easy way to check this is the fact that we know if the insect and the bat are flying at the same speed in the same direction, then there will be no doppler shift. That means if you set f_refl=f_bat, then you should get v_ins=-v_bat from your equation. (The minus sign here when solving means that they are flying in the same direction, since when you set up the signs in your equation you assumed they were flying towards each other.)

Once you fix that sign error, I believe that equation should be correct.

does this make more sense?

 

[kudoushinichi88]

Thanks! I fixed the sign and it became

[tex]

v_{ins}=\frac{f_{bat}v(v+v_{bat})-f_{refl}v(v-v_{bat})}{f_{refl}(v_{bat}-v)-f_{bat}(v+v_{bat})}
[/tex]

which gets v_ins= -v_bat as predicted if f_refl=f_bat

 

[alphysicist]

Thanks! I fixed the sign and it became

[tex]

v_{ins}=\frac{f_{bat}v(v+v_{bat})-f_{refl}v(v-v_{bat})}{f_{refl}(v_{bat}-v)-f_{bat}(v+v_{bat})}
[/tex]

which gets v_ins= -v_bat as predicted if f_refl=f_bat

That all looks right to me.

 

[Delzac]

I have a question here. After the sound from from the bat reflects off the insect, does the sound carry the velocity of the insect too?

So shouldn't the velocity of the sound reflected off the insect be v + v_bat + v_ins?

 

[alphysicist]

I have a question here. After the sound from from the bat reflects off the insect, does the sound carry the velocity of the insect too?

So shouldn't the velocity of the sound reflected off the insect be v + v_bat + v_ins?

I don't believe that is correct. The speed of sound in this problem is constant (v is the speed of sound relative to the air).

If you're referring to the terms in the equation: the terms in the doppler formula numerator is related to the relative speed of the sound and the detector. So for the reflected wave, the numerator is just v+v_bat; that is how fast the sound wave is moving from the bat's point of view.

 

[Delzac]

If the bat is flying at a certain speed and emitting a sound, compared to the bat being stationary and emitting a sound. Shouldn't the former's speed of sound be faster?

And the regarding the reflection of the sound, can i not think of it like how 2 objects elastically collide with each other?

 

[kudoushinichi88]

Speed of sound doesn't change. The moving source/observer will only change the frequency of the observed sound wave.

Sound waves, unlike solid objects, do not collide when they meet each other. They merely interfere.

 

[alphysicist]

If the bat is flying at a certain speed and emitting a sound, compared to the bat being stationary and emitting a sound. Shouldn't the former's speed of sound be faster?

No, emitting sound does not work that way. Of course some things do; for example, a problem might have a dart gun that is shot out of a gun at (say) 2 m/s; if you run while you shoot, the dart will come out faster (relative to the ground) because the dart was already in motion before firing.

But for this problem, the sound that is emitted is not "thrown" by the bat; instead the bat disturbs the air, and then the disturbance is carried away by the air.

And the regarding the reflection of the sound, can i not think of it like how 2 objects elastically collide with each other?

In what way?

 

[Delzac]

Like a tennis ball colliding with a bullet train.

 

[alphysicist]

Like a tennis ball colliding with a bullet train.

There certainly are collisions going on, between air particles and between the air and the object, etc. But how does thinking in terms of the collisions help you find the shifted frequencies?

If you're thinking of something else, just try your idea out on a simple problem and see if it works out. Whether it works out correctly or not, playing around with ideas is the best way to learn.