# Finding peak displacement from RMS sinusoidal acceleration without using frequency

1. Feb 20, 2010

### davidd

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:

$$a_{pk} = 4\pi^2f^2x_{pk}$$

$$a_{rms} = 0.707a_{pk}$$

as...

$$x_{pk} = \frac{a_{rms}}{0.707\pi^2f^24}$$

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

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:

$$x_{pk} = \frac{1}{2}a_{rms}t^21.414$$

$$t = \frac{1}{120} \ s$$ (quarter cycle for travel from 0 to peak)

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

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

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

but I calculated 0.707 from it.

Any ideas what I'm missing?

Last edited: Feb 20, 2010