1
$\begingroup$

Lets say you had a 100g mass bonded to a vibrator which applies a force up and down. It is known that at 100g this mass would oscillate at a maximum pk-pk acceleration of about 5G's, and at a frequency of 100Hz. What would increasing the mass do to the pk-pk acceleration (increase/decrease/nothihng)? What would it do to the frequency (increase/decrease/nothing)?

Below I have a few assumptions.

Assumptions:

  1. The actuator is bonded to ground, and the mass is bonded to the top of the actuator
  2. The actuator only moves in the "up" and "down" direction
  3. The bonding is not elastic (no springiness and no dampening)
  4. The system's "outputs" are taken at "steady state" (not sure if that's the right term for it), basically the vibrator will be on for awhile.

What kind of problem do I have here? I don't think its a simple harmonic oscillator, or a damped simple harmonic oscillator since there is assumed to be no springs or dampers within the system.

So, to continue here is some quick research I've done so far. From what I could find online, "as the mass of a vibrating body increases, its frequency decreases"

Therefore, I can assume that the frequency will likely decrease if I use a 200g mass. Now, I'm a bit stuck on amplitude. Intuitively, I'm assuming that the amplitude should decrease, more mass is harder to push. But, since I'm assuming the vibrator will be on for awhile maybe enough energy will be added to make things work.

I haven't taken any vibrations classes yet so hopefully I'm making a bit of sense. Thanks for any help!

Drawing of problem at hand

$\endgroup$
0
$\begingroup$

Well, considering the vibrations you are talking about are simple harmonic, the frequency and amplitude of its SHM depend on mass in the following manner. $$F=ma=-k.𝑥$$ $$ a=-ω^2𝑥$$ $$ ω^2= k/m $$ where k is the force constant, ω is the angular frequency and 𝑥 is the displacement (Amplitude at extremes). Hence for a fixed value of F (the force applied by the vibrator), acceleration would decrease and so will the frequency. And as F is constant, so is the product of mass and acceleration. Then, $$ ma/k=𝑥 $$ as all the values on left side are constant, the displacement(or amplitude) remains unaffected. The amplitude will not change until some damping or external force is applied.

$\endgroup$
1
  • $\begingroup$ Thanks for your feedback! I apologize if this is a dumb question. But If I was using something like a piezo actuator, would the "spring" in this simple harmonic oscillating system be the inherent stiffness within the actuator? $\endgroup$ Jun 4 at 20:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.