Understanding air particle velocity as cross product freq x disp


by The_Lobster
Tags: cross product, math, particle velocity, sound
The_Lobster
The_Lobster is offline
#1
Apr17-12, 05:20 AM
P: 30
I'm reading a book on microphones and came across the following:

The relation between air particle velocity (u) and particle displacement (x) is given by:
[itex]u(t) = j\omega \times x(t)[/itex]

where [itex] \omega = 2\pi f[/itex] and [itex] x(t) [/itex] is the maximum particle displacement value.
and then it goes off talking about something else...

I feel stupid for asking this, but I don't get how the above equation works? For one, I thought cross products could only be be involving vectors? Aren't all the terms above scalars? Should I treat it as a dot product?

Any help in understanding the above, so I can see how the terms affect each other is greatly appreciated!
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AlephZero
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#2
Apr17-12, 05:55 AM
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I don't think it is a cross product. It looks like the standard equation for velocity of a particle moving in simple harmonic motion.

I don't know why the book used ##\times## as a multiplication sign here.
The_Lobster
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#3
Apr17-12, 05:59 AM
P: 30
Thanks, AlephZero! Typical of me getting thrown off by poor notation...

The_Lobster
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#4
Apr17-12, 06:06 AM
P: 30

Understanding air particle velocity as cross product freq x disp


AlephZero: Are you saying that that equation is pretty much: [itex] v = - A\omega \sin \omega t [/itex]? Does that mean I can consider the "maximum particle displacement" in the first equation, as the amplitude, A?


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