Sound diffraction around obstacles; low and high frequencies

[climbon]

Hi, I am posting here as I can't get a satisfactory answer from google.

Could anyone explain to me why low frequency sounds diffract better than high frequnecy sounds around a corner (eg the wall of a building).

I understand the idea of 'wave diffracts more when opening is around the same size of the wavelength'( doorways for example), but I don't see how this correlates with going around a corner (which doesn't have a 'size of opening' like a doorway has)

Could anyone explain why low frequencies diffract around corners better than high frequencies?

Thank you for any help!

 

[wukunlin]

Think about it this way:

The larger the wavelength is, compared to the size of the gap, the more diffraction you get.

Now, if you only have a corner, you can treat it as one side of a very large gap (so large that you don't need to care about the other side of the gap). When the frequency is lower, the longer the wavelength, the more diffraction you get.

Does that make sense?

 

[wukunlin]

Or, if you prefer a more fundamental description:

Diffraction is a process when a wave with straight wave fronts gets bent after passing through some obstacle. This happens because straight wave fronts are a summation of a bunch of circular wave fronts lined up with each other. (Huygens–Fresnel principle). When passing through an obstacle, some of the circular wave fronts get blocked off, so at the edge of the straight wave that passed through the obstacle, you start to see the circular wave front. The longer the wavelength (or lower the frequency), the more observable the circular wave is, hence "better diffraction."

 

[climbon]

Thanks for your reply!

I understand that now (about the corner behaving as one side of an infinitely large gap). Thanks

I'm still struggling as to why lower frequencies diffract more at a fundamental level. The only reason I can think of is a rather crude explanation by relating it to momentum of light (not sure how this would work for sound?);

The momentum of a photon is = hf/c. So photons with a lower frequency will have a lower momentum; a lower momentum will make it "easier" to deflect.

But it's such a crude way of thinking...maybe I'm clutching at straws lol :)

I'm just really struggling to imagine how a faster vibrating molecule of air diffracts less than a slower vibrating one??

 

[sophiecentaur]

Thanks for your reply!

I understand that now (about the corner behaving as one side of an infinitely large gap). Thanks

I'm still struggling as to why lower frequencies diffract more at a fundamental level. The only reason I can think of is a rather crude explanation by relating it to momentum of light (not sure how this would work for sound?);

The momentum of a photon is = hf/c. So photons with a lower frequency will have a lower momentum; a lower momentum will make it "easier" to deflect.

But it's such a crude way of thinking...maybe I'm clutching at straws lol :)

I'm just really struggling to imagine how a faster vibrating molecule of air diffracts less than a slower vibrating one??

It isn't the frequency that counts - it's the wavelength and the result of the addition of all the possible paths between source and detector that produces nulls and peaks. You can get exactly the same interference pattern with microwaves and ultrasound waves of the same wavelength (say 3cm) where the ratio between the frequencies is around 1000.

Your attempt to explain things in terms of the way the particles move is not valid - unless you consider all the particles in the region of the experiment (e.g. the room) and that is best done using the wave model.

Have you looked up interference and diffraction on Wiki?