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Finding peak displacement from RMS sinusoidal acceleration without using frequency

  1. Feb 20, 2010 #1
    I am looking at sinusoidal motion (shaker tables) and am trying to calculate peak displacement given RMS acceleration.

    I can easily find the peak displacement using:

    [tex]a_{pk} = 4\pi^2f^2x_{pk}[/tex]

    [tex]a_{rms} = 0.707a_{pk}[/tex]


    [tex]x_{pk} = \frac{a_{rms}}{0.707\pi^2f^24} [/tex]

    My question is: isn't there a way to find the peak acceleration without using frequency and [tex]\pi[/tex] ?

    I am stuck thinking that there is since RMS voltage and current can be used to find power without the frequency being involved. I initially thought this would work:

    [tex]x_{pk} = \frac{1}{2}a_{rms}t^21.414 [/tex]

    [tex]t = \frac{1}{120} \ s[/tex] (quarter cycle for travel from 0 to peak)

    The values I've been using to work through this are: [tex]a_{pk} = 29.4 \frac{m}{s^2}, \ \ \ a_{rms} = 20.8 \frac{m}{s^2}, \ \ \ f = 30 Hz, \ \ \ x_{pk} = 0.825 mm [/tex]

    I even verified that RMS acceleration is the same for a quarter cycle as it is for a full cycle by doing this calculation:

    [tex]a_{rms} = \sqrt{\frac{1}{\frac{3}{120} - \frac{1}{120}} \int^{\frac{3}{120}}_{\frac{1}{120}} sin(2\pi30t)^2 dt [/tex]

    but I calculated 0.707 from it.

    Any ideas what I'm missing?
    Last edited: Feb 20, 2010
  2. jcsd
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