Doppler Effect Question: Flying Bat Echolocating a Mosquito

[pot]

Homework Statement

Assume the speed of sound to be 340 m/s. A bat flying at 1.05 m/s chirps at a frequency of 10.3 kHz. At the same time, a mosquito is flying away from the bat at 1.00 m/s.

1. What is the frequency of the sound that returns to the bat?

2. What would your answer be if the bat were not moving, but hanging from a tree?

3. If bats can detect a change in frequency as low as 1.00 Hz, can the bat detect the mosquito? Justify your answer.

Relevant Equations

f' = f [(v +/- vo) / (v +/- vs)]
kHz x1000 -> Hz

Hi,

Would 1) be f' = (10,300)[(340 - 1.00) / (340 - 1.05)] = 10,302 Hz
2) f' = (10,300)[(340 - 1.00) / (340)] = 10,270 Hz
3) Yes? As the frequency changes more and less than 1.00 Hz.

I am unsure and would appreciate any help. Thanks!

 

[haruspex]

Homework Statement: Assume the speed of sound to be 340 m/s. A bat flying at 1.05 m/s chirps at a frequency of 10.3 kHz. At the same time, a mosquito is flying away from the bat at 1.00 m/s.

1. What is the frequency of the sound that returns to the bat?

2. What would your answer be if the bat were not moving, but hanging from a tree?

3. If bats can detect a change in frequency as low as 1.00 Hz, can the bat detect the mosquito? Justify your answer.
Relevant Equations: f' = (v +/- vo) / (v +/- vs)
kHz x1000 -> Hz

Hi,

Would 1) be f' = (10,300)[(340 - 1.00) / (340 - 1.05)] = 10,302 Hz
2) f' = (10,300)[(340 - 1.00) / (340)] = 10,270 Hz
3) Yes? As the frequency changes more and less than 1.00 Hz.

I am unsure and would appreciate any help. Thanks!

Your relevant equation is missing an f on the RHS.
Exactly what circumstance does that equation apply to? Does this match the question?

 

[pot]

Oops, sorry about that. I still included the frequency value in my calculations. I fixed it now.

I thought the Doppler shift equation would be used when calculating frequencies that are effected by other speeds. Do you mean I have chosen the wrong +/-?

Rereading the question now, I think 1) would be f' = (10,300)[(340 - 1.00) / (340 + 1.05)] = 10,238 Hz

As the bat moves away from where it was initially and the mosquito moves away as well. And 2) would still be the same f' = (10,300)[(340 - 1.00) / (340)] = 10,270 Hz

Your relevant equation is missing an f on the RHS.
Exactly what circumstance does that equation apply to? Does this match the question?

 

[vela]

3) Yes? As the frequency changes more and less than 1.00 Hz.

Which is it, more or less? It can't be both at the same time.

I thought the Doppler shift equation would be used when calculating frequencies that are effected by other speeds. Do you mean I have chosen the wrong +/-?

No, he's asking if you have analyzed the problem correctly. The Doppler shift formula applies to a specific situation, and you're not using it right.

 

[pot]

Which is it, more or less? It can't be both at the same time.


No, he's asking if you have analyzed the problem correctly. The Doppler shift formula applies to a specific situation, and you're not using it right.

For 3) Yes, as the bat can detect a change in frequency as low as 1.00 Hz, and the frequency in both 1) and 2) change by more than 1 Hz (from 10,300 Hz to less than 10,299 Hz).

And about the Doppler shift formula, I'm confused. Can you explain how I'm not using it right?

 

[haruspex]

For 3) Yes, as the bat can detect a change in frequency as low as 1.00 Hz, and the frequency in both 1) and 2) change by more than 1 Hz (from 10,300 Hz to less than 10,299 Hz).

And about the Doppler shift formula, I'm confused. Can you explain how I'm not using it right?

The formula gives the received frequency for a sound sent from a moving source directly to a moving receiver. Who is the receiver in the question?

 

[Orodruin]

To be a bit more explicit without revealing too much … the Doppler formula refers to a single signal travelling from a source to an observer. This is not the scenario here: You have a signal emitted from the bat, reflected on the mosquito, and finally observed by the bat.

Ask yourself the following questions:
1. What frequency is observed by the mosquito for the original signal?
2. What will be the frequency of the reflected signal as measured ny its source (the mosquito)
3. What will be the frequency of the reflected signal observed by the bat?

 

[pot]

To be a bit more explicit without revealing too much … the Doppler formula refers to a single signal travelling from a source to an observer. This is not the scenario here: You have a signal emitted from the bat, reflected on the mosquito, and finally observed by the bat.

Ask yourself the following questions:
1. What frequency is observed by the mosquito for the original signal?
2. What will be the frequency of the reflected signal as measured ny its source (the mosquito)
3. What will be the frequency of the reflected signal observed by the bat?

So for 1) I would have to use the formula twice? Like this:
f(mos. hears) = (10,300)[(340 - 1)/(340 - 1.05)] = 10,302 Hz
Then: f(bat hears) = (10,301)[(340 + 1.05)/(340 + 1)] = 10,303 Hz

2) f(bat hears) = (10,301)[340/(340 + 1)] = 10,271 Hz

 

[haruspex]

So for 1) I would have to use the formula twice? Like this:
f(mos. hears) = (10,300)[(340 - 1)/(340 - 1.05)] = 10,302 Hz
Then: f(bat hears) = (10,301)[(340 + 1.05)/(340 + 1)] = 10,303 Hz

It looks bit strange the way 10,302 became 10,301, but I assume you actually calculated to greater precision, getting 10,301.5 as the intermediate result. The final number is correct.

2) f(bat hears) = (10,301)[340/(340 + 1)] = 10,271 Hz

It's the same set-up, just different speeds. You still need to consider both directions.

 

[pot]

Oh, I see! So for 2) I would have to use the formula twice again so it would be f(mos. hears) = (10,300)[(340 - 1)/340] = 10,270 Hz and then: f(bat still) = (10,270)[340 / (340+1)] = 10,240 Hz

 

[haruspex]

Oh, I see! So for 2) I would have to use the formula twice again so it would be f(mos. hears) = (10,300)[(340 - 1)/340] = 10,270 Hz and then: f(bat still) = (10,270)[340 / (340+1)] = 10,240 Hz

Yes.
You can make the numerical calculations a little simpler by noting that the velocities are small compared with the speed of sound, allowing approximations:
$f'=f\frac{c-v_m}{c-v_b}\frac{c+v_b}{c+v_m}=f\frac{1-v_m/c}{1-v_b/c}\frac{1+v_b/c}{1+v_m/c}\approx f(1-v_m/c+v_b/c +v_b/c-v_m/c)$
$=f(1+2(v_b-v_m)/c)$

 

[pot]

That makes sense. Thank you so much for the help!

 

[PeroK]

That makes sense. Thank you so much for the help!

Elementary physics is often taught as "plug and chug", where you are expected simply to plug numbers into an equation without thinking much about the problem.

The opposite of plug and chug is post #7, by @Orodruin, where the physical problem is analysed.

As you tackle more advanced problems, analysing them becomes more important than plugging numbers into an equation.