Эротические рассказы

Vibroacoustic Simulation. Alexander PeifferЧитать онлайн книгу.

Vibroacoustic Simulation - Alexander Peiffer


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structures and thus, how they exchange acoustic energy.

      2.4.1 Monopoles – Spherical Sources

       x equals r sine theta cosine phi (2.62a)

       y equals r sine theta sine phi (2.62b)

       z equals r cosine theta (2.62c)

      Using this coordinate system and neglecting the angular components the Laplace operator Δ reads as

       normal upper Delta equals StartFraction 1 Over r squared EndFraction StartFraction partial-differential Over partial-differential r EndFraction left-parenthesis r squared StartFraction partial-differential Over partial-differential r EndFraction right-parenthesis equals StartFraction 2 Over r EndFraction StartFraction partial-differential Over partial-differential r EndFraction plus StartFraction partial-differential squared Over partial-differential r squared EndFraction (2.63)

      The wave equation for the velocity potential (2.30) becomes

       left-parenthesis StartFraction 1 Over c 0 squared EndFraction StartFraction partial-differential squared Over partial-differential t squared EndFraction minus StartFraction 2 Over r EndFraction StartFraction partial-differential Over partial-differential r EndFraction minus StartFraction partial-differential squared Over partial-differential r squared EndFraction right-parenthesis normal upper Phi equals 0 (2.64)

      The two right terms can be written in a different form using rΦ as argument

       left-parenthesis StartFraction 1 Over c 0 squared EndFraction StartFraction partial-differential squared Over partial-differential t squared EndFraction minus StartFraction 1 Over r EndFraction StartFraction partial-differential squared Over partial-differential r squared EndFraction right-parenthesis left-parenthesis r normal upper Phi right-parenthesis equals 0 period (2.65)

      Equation (2.30) is the one-dimensional wave equation for the argument rΦ, so we can use the D’Alambert solution

      The first term represents an outgoing wave travelling away from the source, the second an incoming wave travelling to the source. As we are interested in sound being emitted from the source we consider the outgoing harmonic solution with complex amplitude A

      Consider a pulsating sphere of radius R in the centre with normal surface velocity vR. With the velocity potential the radial velocity can be easily derived from the solution (2.67):

      Substituting Equation (2.67) into (2.68) and solving for A gives

       bold-italic upper A equals minus bold-italic v Subscript upper R Baseline StartFraction upper R squared Over 1 minus j k upper R EndFraction e Superscript j k a (2.69)

      Hence,

       normal upper Phi left-parenthesis r comma t right-parenthesis equals minus StartFraction bold-italic v Subscript upper R Baseline Over r EndFraction StartFraction upper R squared Over 1 minus j k upper R EndFraction e Superscript j left-bracket omega t minus k left-parenthesis r minus upper R right-parenthesis right-bracket (2.70)

      The strength Q(t) of the source is defined by the volume flow rate. This is the surface of the sphere times normal velocity vR

       upper Q left-parenthesis t right-parenthesis equals ModifyingAbove upper V With dot equals 4 pi upper R squared v Subscript upper R Baseline left-parenthesis t right-parenthesis (2.71)

      With the harmonic source strength

       StartLayout 1st Row 1st Column bold-italic upper Q left-parenthesis t right-parenthesis 2nd Column equals 4 pi upper R squared bold-italic v Subscript upper R Baseline e Superscript j omega t Baseline 3rd Column bold-italic upper Q left-parenthesis omega right-parenthesis 4th Column equals 4 pi upper R squared bold-italic v Subscript upper R EndLayout (2.72)

      the spherical wave solution is

      and

       bold-italic v Subscript r Baseline left-parenthesis r comma omega right-parenthesis equals minus left-parenthesis StartFraction <hr><noindex><a href=Скачать книгу

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