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Mechanical waves

A

Adam

Jan 1, 1970
0
Hi,

I have a question about mechanical waves (sound and ultrasonic waves
for instance).
I am familiar with wavelength formula: lambda=C/F
But I am not able to understand why a sound wave (10 KHz for instance)
which has a LONGER wavelength than an ultrasonic wave can pass through
a barrier like a thin wall while an ultrasonic wave (50 KHz for
instance) will reflect from it??

I know that my question is a physical question not an electronically
question but I noticed that problem when I was working with different
speakers.

Thanks for any advice.
 
S

Stanislaw Flatto

Jan 1, 1970
0
Adam said:
Hi,

I have a question about mechanical waves (sound and ultrasonic waves
for instance).
I am familiar with wavelength formula: lambda=C/F
But I am not able to understand why a sound wave (10 KHz for instance)
which has a LONGER wavelength than an ultrasonic wave can pass through
a barrier like a thin wall while an ultrasonic wave (50 KHz for
instance) will reflect from it??

I know that my question is a physical question not an electronically
question but I noticed that problem when I was working with different
speakers.

Thanks for any advice.
As both items are of the same nature it boils down to attenuation
properties of the barrier. In mechanical (acustic) matters there is NO
such thing as perfect reflection. Make your measurements sensitive few
orders of magnitude more and the results will be VERY instructive.

Good luck.

Stanislaw.
 
C

Calvin Guan

Jan 1, 1970
0
I don't know much about sound wave but all waves can be described by a set
of partial differential equations (PDE). when wave passes a boundary between
2 medium (air and wall), the PDE must be satisfied at the boundary. this
usually yields 2 products -- the transmitted wave, and the reflected wave.
This is relatively easy to imagine if you think of a traveling EM wave
entering a dielectric medium from free space, at the boundary, the E&M
fields must satisfied the Maxwell's Equations. The result is frequency and
medium dependent. For some condition, the transmitted wave is much greater
than the reflected (S21>>S11), for other conditions, it's the other way
(S11>>S21).

By solving your PDE, it shoul tell how much energy got reflected vs
transmitted.
hth,
Calvin
 
J

John Larkin

Jan 1, 1970
0
Hi,

I have a question about mechanical waves (sound and ultrasonic waves
for instance).
I am familiar with wavelength formula: lambda=C/F
But I am not able to understand why a sound wave (10 KHz for instance)
which has a LONGER wavelength than an ultrasonic wave can pass through
a barrier like a thin wall while an ultrasonic wave (50 KHz for
instance) will reflect from it??

I know that my question is a physical question not an electronically
question but I noticed that problem when I was working with different
speakers.

Thanks for any advice.

A thin, flexible wall, exposed to low-frequency sound, will move with
the pressure waves and act like a piston on the far side, so will be
almost transparent to sound. An infinitely stiff wall of any
thickness, or a wall of very high density, will reflect most of the
sound.

At high frequencies, as the mass of the wall becomes harder to move at
the sound frequency, the wall becomes essentially immobile and classic
"optical" reflection/transmission rules start to apply. Since the
wall's density is generally a lot higher than air's, in "optical"
situations most of the sound will be reflected.

That's why you hear mostly the bass of your neighbor's terrible, loud
music. It's shaking your walls.

In optics, the incoming wave doesn't cause a wall to move, so the
behavior is simpler.

There's probably some formal math, probably googl'able, on sound
transmission through thin, flexible walls.

John
 
J

john jardine

Jan 1, 1970
0
Adam said:
Hi,

I have a question about mechanical waves (sound and ultrasonic waves
for instance).
I am familiar with wavelength formula: lambda=C/F
But I am not able to understand why a sound wave (10 KHz for instance)
which has a LONGER wavelength than an ultrasonic wave can pass through
a barrier like a thin wall while an ultrasonic wave (50 KHz for
instance) will reflect from it??

I know that my question is a physical question not an electronically
question but I noticed that problem when I was working with different
speakers.

Thanks for any advice.

Have a look at http://www.falstad.com/mathphysics.html
There's a whole range of magnificent physics demonstrations done as Java
'applets'. Enormous fun!.
Play with the '2D ripple tank' and use your mouse to add a couple of walls
and slots etc. It's then clear as to what happens to the reflections when
the frequency is taken up and down.
Seems to me that the long waves can't cleanly reflect, as they are simple
too diffuse wrt the obstacles. I.e they've gone round/slid along the
obstacle before the wave reverses.
 
P

Phil Allison

Jan 1, 1970
0
"Adam"
I am familiar with wavelength formula: lambda=C/F
But I am not able to understand why a sound wave (10 KHz for instance)
which has a LONGER wavelength than an ultrasonic wave can pass through
a barrier like a thin wall while an ultrasonic wave (50 KHz for
instance) will reflect from it??


** Your facts are cockeyed.

For 10kHz to pass through a wall, it would have to be made from thin paper.

All solid walls refect sound pressure waves and only to the extent they can
be made to vibrate (at the frequency concerned ) will that sound wave pass
to the opposite side.

A solid, smooth wall reflects all sound and supersonic frequencies.




....... Phil
 
F

Fred Bartoli

Jan 1, 1970
0
Le Wed, 26 Sep 2007 00:40:21 +0100, john jardine a écrit:
Have a look at http://www.falstad.com/mathphysics.html There's a whole
range of magnificent physics demonstrations done as Java 'applets'.
Enormous fun!.
Play with the '2D ripple tank' and use your mouse to add a couple of
walls and slots etc. It's then clear as to what happens to the
reflections when the frequency is taken up and down.
Seems to me that the long waves can't cleanly reflect, as they are
simple too diffuse wrt the obstacles. I.e they've gone round/slid along
the obstacle before the wave reverses.

Great link!
Thanks.
 
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