Sound Waves/Doppler

[MedPR]

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What frequency would a bat use to locate an insect 1cm wide which is 10m away? Assume speed of sound is 340m/s. The passage also says that the sound must encounter an obstacle which is larger than the wavelength of the sound in order for the bat to detect the insect.

A. Any frequency less than 34kHz
B. Any frequency greater than 34kHz
Others are irrelevent.


Explanation:

B. According to paragraph 1, the wavelength must be shorter than the target insect. Thus the frequency must be greater than f=v/linsect = 343000 Hz = 34.3 kHz

How does this prove that the wavelength of the sound is shorter than the target insect? v=frequency*lambda

343=343000/lambda

Lambda=.001 m = 1cm?

I kind of figured this out as I typed it, but I figured I would put it up anyway just in case someone else needed it 🙂 Also, is the bold even correct?

 

[milski]

Sort of correct, except that you made it work with an even number of math errors. 🙂

34.3 kHz = 34300 Hz
lambda=v/f=343 / 34000 = 0.01 m = 1 cm
Higher frequencies will mean smaller lambda, so you want f>34.3 kHz

 

[SaintJude]

The bat (source) will push out a sound wave to the insect (the source) that will then reflect back from the target to the bat. So the bat will register a lower perceived frequency, because the insect and bat are approaching each other. Thus, to get an accurate reading the bat will have actually projected a larger frequency than it perceives the reflection has sent.

 

[MedPR]

Sort of correct, except that you made it work with an even number of math errors. 🙂

34.3 kHz = 34300 Hz
lambda=v/f=343 / 34000 = 0.01 m = 1 cm
Higher frequencies will mean smaller lambda, so you want f>34.3 kHz

Right, thanks 🙂 I was rushing to type it out before class started.

The bat (source) will push out a sound wave to the insect (the source) that will then reflect back from the target to the bat. So the bat will register a lower perceived frequency, because the insect and bat are approaching each other. Thus, to get an accurate reading the bat will have actually projected a larger frequency than it perceives the reflection has sent.

The problem didn't say anything about whether or not the bat/insect were approaching each other. The concept being tested in the question is that the wavelength emitted by the source (bat) must be smaller than the size of the object it is to be bounced off of (the insect). Though, in a case where the source and observer are approaching, you are correct that the frequency increases, thus decreasing the wavelength.

 

[SaintJude]

Oh, sorry, I didn't realize that this was a passage-based question and they had told you that lambda must be smaller. I had a similar discrete questions with a bat (I guess doppler & bats are very popular), so I just applied that scenario.

 

[MedPR]

Oh, sorry, I didn't realize that this was a passage-based question and they had told you that lambda must be smaller. I had a similar discrete questions with a bat (I guess doppler & bats are very popular), so I just applied that scenario.

semi-hat trick without even knowing it 🙂