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# Hard physics question!! - AS Watch

1. ll
2. (Original post by Timlong365)
Please help me with this edexcel physics question. It seems impossible. I can't even understand the mark scheme, nor can my teacher.

The frequency of sound in an oilbird's click is 2.0 kHz and a click is 2.0 ms in duration.
Determine whether these clicks can be used to locate obstacles at a distance of 50 cm.
Speed of sound = 330 m/s (5 marks)
I've attempted this question but I'm not sure if what i've concluded is correct. Using the distance time equation: I've calculated that the wave will travel 66cm within a time interval of 2ms at a speed of . Then using the wave speed formula . I calculated the wavelength as being 16.5cm. Those parts that I have calculated I am confident with, now onto the second parts. The duration of the wave being 2ms means that there will be two separate wavefronts, and these wavefronts will have a distance separating them of . I calculated this distance by multiplying the wavelength () by the duration (). Now, coincidentally, when you divide the wavelength () by this distance of () you get . That's where I've got to right now and I'm not sure if any of it is relevant, would you mind posting the markscheme so I could also have a look at it. Thanks
3. (Original post by mrtomsyw)
I've attempted this question but I'm not sure if what i've concluded is correct. Using the distance time equation: I've calculated that the wave will travel 66cm within a time interval of 2ms at a speed of . Then using the wave speed formula . I calculated the wavelength as being 16.5cm. Those parts that I have calculated I am confident with, now onto the second parts. The duration of the wave being 2ms means that there will be two separate wavefronts, and these wavefronts will have a distance separating them of . I calculated this distance by multiplying the wavelength () by the duration (). Now, coincidentally, when you divide the wavelength () by this distance of () you get . That's where I've got to right now and I'm not sure if any of it is relevant, would you mind posting the markscheme so I could also have a look at it. Thanks
Thanks for answering! This is the mark scheme:
Use of s = v5/2 (to find pulse length/2 as minimum distance to object)
s = 0.33m
use of v=f*lamda
lamda = 0.17m
half pulse length needs to be less thn distance and half wavelength needs to be less than distance; both are, so clicks are suitable.
Do you understand why wavelength is relevent in this question? :/
4. (Original post by mrtomsyw)
I've attempted this question but I'm not sure if what i've concluded is correct. Using the distance time equation: I've calculated that the wave will travel 66cm within a time interval of 2ms at a speed of . Then using the wave speed formula . I calculated the wavelength as being 16.5cm. Those parts that I have calculated I am confident with, now onto the second parts. The duration of the wave being 2ms means that there will be two separate wavefronts, and these wavefronts will have a distance separating them of . I calculated this distance by multiplying the wavelength () by the duration (). Now, coincidentally, when you divide the wavelength () by this distance of () you get . That's where I've got to right now and I'm not sure if any of it is relevant, would you mind posting the markscheme so I could also have a look at it. Thanks
This is all about echo location.

The bird both chirps and then listens for the return.

In order to locate the object at a distance of 50cm, the chirp must be of sufficient frequency and of enough repetition that the bird is able to hear the signal clearly.

That means several cycles must be heard after the transmitted sound stops otherwise the two will overlap and confuse the bird.

This places constraints on the minimum distance that can be detected:

The bird chirp is 2mS duration. Which means 4 whole cycles @ 2kHz are produced in that time.
The wavelength of 16.5cm means that a maximum of (2x50)/16.5 > 6 full wave cycles will be detected before the transmitted-received sound overlaps when reflected off the obstacle at 50cm.

Conclusion: Yes the bird can detect obstacles at 50cm.
5. (Original post by uberteknik)
This is all about echo location.

The bird both chirps and then listens for the return.

In order to locate the object at a distance of 50cm, the chirp must be of sufficient frequency and of enough repetition that the bird is able to hear the signal clearly.

That means several cycles must be heard after the transmitted sound stops otherwise the two will overlap and confuse the bird.

This places constraints on the minimum distance that can be detected:

The bird chirp is 2mS duration. Which means 4 whole cycles @ 2kHz are produced in that time.
The wavelength of 16.5cm means that a maximum of (2x50)/16.5 > 6 full wave cycles will be detected before the transmitted-received sound overlaps when reflected off the obstacle at 50cm.

Conclusion: Yes the bird can detect obstacles at 50cm.
So all you had to do for this question was calculate wavelength and the number of waves emitted during a single click and show if enough full wave cycles will get detected? Does this mean that the minimum for an obstacle to be located would be around 24.75cm as 49.5/16.5 = 3 whole wave cycles?
Why does the mark scheme say "half wavelength needs to be less than distance"?
6. (Original post by Timlong365)
Please help me with this edexcel physics question. It seems impossible. I can't even understand the mark scheme, nor can my teacher.

The frequency of sound in an oilbird's click is 2.0 kHz and a click is 2.0 ms in duration.
Determine whether these clicks can be used to locate obstacles at a distance of 50 cm.
Speed of sound = 330 m/s (5 marks)
Is it really necessary to post the same thread twice??
7. (Original post by emiloujess)
Is it really necessary to post the same thread twice??
I'm very sorry, but it wasn't getting any answers in the first one I posted and this is a very important exam question and my actual physics exam is in around a week so I was very desperate to have it answered.
8. (Original post by Timlong365)
So all you had to do for this question was calculate wavelength and the number of waves emitted during a single click and show if enough full wave cycles will get detected? Does this mean that the minimum for an obstacle to be located would be around 24.75cm as 49.5/16.5 = 3 whole wave cycles?
Why does the mark scheme say "half wavelength needs to be less than distance"?
It's not a very good question (on the examiners part) because we don't know the biology and ability of the bird. Assumptions are therefore necessary which is not good in an exam as it introduces ambiguity.

Because the chirp is 4 cycles duration (which must all be completed because that's what the bird chirps) then we have to make an assumption that the bird knows when it stops chirping and starts listening for the echo. The only unambiguous way to do that is when the bird hears the whole of the chirp (all four cycles) after it stops chirping and starts listening for the echo.

That means the minimum time for the whole sound to make the return journey is just greater than 2mS before the bird cannot tell the difference between the transmitted sound and the echo sound.

i.e. 2mS x 330m/s = 66cm.

However this is the return distance the whole waveform can travel, so the minimum distance to the object is half that at 33cm.
9. (Original post by Timlong365)
I'm very sorry, but it wasn't getting any answers in the first one I posted and this is a very important exam question and my actual physics exam is in around a week so I was very desperate to have it answered.
so bump the original thread, not make a new one!! Btw there IS an answer in the other thread sooo...

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