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How do I go about finding/calculating the specifications of an accelerometer that can measure vibrations upto 2kHz and deflections upto 2mm?

EDIT: After calculating and obtaining 16000g's worth of acceleration, I consulted my mentor and it seems like we might not experience 2mm of deflections at 2kHz. However, now the problem becomes how to estimate what deflections we will see at that frequency if we use PL 140 actuators from PI and place them on the panel?

https://www.piceramic.com/en/products/piezoceramic-actuators/bender-actuators/pl112-pl140-picma-bender-103000/

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    \$\begingroup\$ It seems to me that you have defined them already a lot better than many questions that have been asked on the site. Edit your question to explain the application a bit (single axis, bandwidth, etc.) and work out what acceleration you will need at that frequency and deflection. \$\endgroup\$ – Transistor Jul 7 '17 at 15:58
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    \$\begingroup\$ First you'd need to calculate the expected acceleration from the deflection, frequency, and expected wave shape. Then go to a parametric search, both by acceleration range, and reading frequency if you want to dry to do a reasonable job of sampling the acceleration waveform. \$\endgroup\$ – Chris Stratton Jul 7 '17 at 15:58
  • \$\begingroup\$ oh, and in practice it's also often important what kind of output signal your sensor has – ie. whether it modulates a voltage or a current, or gives digital readings, or... \$\endgroup\$ – Marcus Müller Jul 7 '17 at 16:05
  • \$\begingroup\$ are you asking us to calculate the peak acceleration due to a 2mm amplitude at 2kHz? \$\endgroup\$ – Neil_UK Jul 7 '17 at 16:07
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    \$\begingroup\$ The linked web page for that piezo actuator lists its dimensions as 45 mm x 11 mm and a "blocking force" of +/- 0.5 N. If this means that 0.5 N is enough to stop it then it seems to me that an accelerometer may load it considerably as will whatever it is driving. That +/-1000 µm displacement probably only occurs with no load. (I know close to nothing about these types of devices!) \$\endgroup\$ – Transistor Jul 8 '17 at 1:31
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Accelerometers are specified by the maximum accelerations they can measure. You therefore have to calculate what acceleration is represented by moving 2 mm at 2 KHz.

From your description, 2 mm is the peak to peak magnitude of the 2 kHz oscillation. Acceleration is the second derivative of displacement. The displacement is (1 mm)sin(2Πt 2000/s). Take the second derivative, which will be another sine. The peak of that is the ±magnitude the accelerometer needs to be able to measure. Acclerometers are often specified in g instead of m/s2 or the like. 1 g = 9.8 m/s2.

Added

Now I have time to show the calculations. The equation for displacement is:

    Disp = (1 mm)sin(2Πt 2000/s)

First derivative:

    Speed = (12.57x103 mm/s)cos(2Πt 2000/s)

Second derivative:

    Acc = -(158x106 mm/s2)sin(2Πt 2000/s)

The accelleration range is therefore ±158x106 mm/s2 = ±158x103 m/s2.

    ±(158x103 m/s2)/(9.8 m/s2g) = ±16.1x103 g.

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  • \$\begingroup\$ I get an extremely large amount of g's. Just the magnitude is 315827.34. It's about 32227.28g. I don't think that's correct. \$\endgroup\$ – Yukti Kathuria Jul 7 '17 at 17:00
  • \$\begingroup\$ @Yuk: I just realized I made a mistake since it's 2 mm peak to peak, so 1 mm peak (now fixed above). Still that only accounts for a factor of 2. I don't have time now to go thru the numbers myself. \$\endgroup\$ – Olin Lathrop Jul 7 '17 at 17:04
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    \$\begingroup\$ The 16000g are correct. \$\endgroup\$ – sweber Jul 7 '17 at 17:36

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