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Mechanical waves

Discussion in 'Electronic Design' started by Adam, Sep 25, 2007.

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  1. Adam

    Adam Guest


    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

    Thanks for any advice.
  2. 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.

  3. Calvin Guan

    Calvin Guan Guest

    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

    By solving your PDE, it shoul tell how much energy got reflected vs
  4. John Larkin

    John Larkin Guest

    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

    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.

  5. john jardine

    john jardine Guest

    Have a look at
    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.
  6. Phil Allison

    Phil Allison Guest


    ** 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
  7. Fred Bartoli

    Fred Bartoli Guest

    Le Wed, 26 Sep 2007 00:40:21 +0100, john jardine a écrit:
    Great link!
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