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A teacher stands well back from an outside doorway that is 95 cm wide and blows a frequency of 600 Hz. The door faces north. Assume the speed of sound is 340 m/s and that there are no reflections. At what angles (from the line normal to the doorway) is it not possible to hear the whistle outside on the playground?

I've already got the wavelength to be 56.67 cm and I know that i should be applying the double slit equation for destructive interference, but I don't know how to apply that equation for this situation.


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3 Answers 3

It isn't the double slits equation you need to use, for the very good reason that there is only one slit. Instead you need to know the angular resolution due to diffraction of the sound by the doorway. The phrase "A teacher stands well back" suggests you are intended to assume the sound has negligible angular divergence when it reaches the door, and the angular divergence is just due to diffraction.

You might want to have a look at the Wikipedia article on angular resolution for some clues.

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and don't forget you'd need to know in detailed how good the teacher's hearing is, both in frequency range and volume. Maybe he's almost deaf and can't hear anything even when standing right in front of the speaker :) –  jwenting Feb 6 '13 at 9:43
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The Wikipedia page on Single Slit diffraction http://en.wikipedia.org/wiki/Diffraction#Single-slit_diffraction gives a good explanation of how to calculate these minima for that case that the slit is larger than the wavelength of the monochromatic source, which in your case it is.

In essence you can imagine there being a series of point sources along the length of the slit all spreading out. Picking any 2 of these sources, there will be destructive interference when the path difference between them is $n \lambda / 2 $. Using some trigonometry it can then be shown that the angles $\theta$ for which the minima will occur is given by:

$sin(\theta_{n}) = (n \lambda / d)$

where $d$ is the width of the single slit, $\lambda$ is the wavelength and $n$ is an integer giving the $n^{th}$ minima from the centre, e.g. $n = 1$ is the first minima.

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You are supposed to use the single slit diffraction formula here which is: $$a\times\sin\theta = m\times\lambda $$ $$\theta = arcsin(m\times\lambda/a)$$

This formula gives you the 'dark fringes' that would occur under certain angles for $m$ element of Z \ {0}.

To find the exact angles (the answer to your question), fill in the values for different values of m ;-)

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Thanks Crazy buddy for the latex fix :D –  Spyral Feb 6 '13 at 10:17
Please see our homework policy. We only give partial answers to HW questions. I shall temporary delete your answer for now. If you edit it so that it only provides hints (and not a complete solution), I'll undelete it (flag it to get my attention or ping me in chat) –  Manishearth Feb 6 '13 at 11:53
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