# Calculate Natural frequency of a vibrator

1. Dec 10, 2012

### Pivskid

I have a construction of a vibrator (industrial use).
I should be able to calculate the natural frequency, by means of Hooke's law.
Since I have a physical device I already have a "real world" measurement of the Natural frequency.
The problem is: How to calculate it?
I have tried to do some equivalent work, and a calculation that is far of the measured value.
You could say "why calculate when you have the measured value".
The vibrator at hand now, is only a model.
I need to be able to calculate the NF fairly accurate, in order not to demolise the building where the real one is installed )

Observe that the 2 mass'es are placed in an angle of 30° from horizontal.
That is..mass2 is higher than mass1.
All springs are equal, eventhough the drawing might insinuate otherwise.

Last edited: Dec 10, 2012
2. Dec 10, 2012

### Staff: Mentor

What does the central object do?
Are the masses free to move in 2 (or even 3) dimensions? This would make an analytic approach tricky.

A naive ansatz:
Assume that 4kg/mm really means 40N/mm (kg/m is not a meaningful unit for springs).
Replace the 6 springs by an effective spring of (4+1/2)*40N/mm = 180N/mm. Use symmetry to reduce the problem to 1 mass with an effectice spring of D=360N/mm, attached to a fixed end. This gives $f=\frac{1}{2\pi} \sqrt{\frac{D}{m}} = 27.6 Hz$.
Working with g=9.81m/s^2 instead of g=10m/s^2 reduced D by 2% and f by 1%, so the result would be ~27.3Hz. Both values are quite close to the measured frequency.

3. Dec 11, 2012

### Pivskid

The central object, I assume you mean the eccentric, it drives the vibrator.
It is a motor with an eccentric attaached to the shaft.
The 2 masses are moving in one direction (the 30° from horizontal).
I understand how you get the 4kg/mm to 40N/mm, since it is essentially the same.
But then I'm lost.
You replace the 6 springs into 4+½, where does the ½ come from ?
Then you reduces the 2 masses to 1 by symmetry.. how? (mass1+mass2 ?)
And you end up with an equivalent spring twice as high 360N/mm ?

4. Dec 11, 2012

### Staff: Mentor

4 springs are connecting both masses, they can be added. The two middle springs can be replaced with an effective spring of 20N/mm. If you use 20N of force, you compress both springs by 1/2 mm and therefore the total length reduces by 1mm.

In the same way, a 180N/mm-spring can be replaced by a 360N/mm-spring if I just look at the compression on one side. This is the other direction.

5. Dec 11, 2012

### Pivskid

ohh you see spring K5 and K6 as being in series.
Well there might be a catch to that, since the eccentric is by the propelling motor is attached to the frame.
I still do not get what you did to the mass m in the equation (=240N ?)

I cannot get your formula to result in 27,6Hz (

6. Dec 11, 2012

### Staff: Mentor

That is no problem, you can consider just one side all the time if that is easier to visualize. There, one spring does not matter at all.

D=360N/mm, m=12kg.
Did you consider the mm <-> m conversion?

7. Dec 13, 2012

### Pivskid

Yes, I let the 360N/mm -> 360kN/m
I now understand you set the mass to 12kg not 120N, in the equation, as I expected.
Ít surprises me a little. Why in kg and not N ?