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# Natural frequency of a damped system, oscillated with a sinusoidal displacement

## Natural frequency of a damped system, oscillated with a sinusoidal displacement

(OP)
Hi!

I have the following system:

x is am imposed motion and y is the output that I obtain.

I would like to calculate the natural frequency of the system. My intuition says that it is equal to sqrt((K+k)/m), but I am not sure how to derive this value from the equation of motion of the system.

Could you help me with this?

### RE: Natural frequency of a damped system, oscillated with a sinusoidal displacement

Your intuition is correct enough when the damping is small.  This is because your two springs are, in effect, "in parallel", and so their stiffnesses can simply be added.

For finite, but still sub-critical, damping you have to multiply the sqrt(k/m) value by sqrt(1-e^2) where e is the damping ratio.  As is self-evident from the mathematical form of this factor, the result is valid only for damping ratios less than unity, ie for underdamped systems.

### RE: Natural frequency of a damped system, oscillated with a sinusoidal displacement

(OP)
Hi!

Yes, actually I forgot to say that I assume underdamped system of negligible damping factor for my intuition :). So, it is as you say.

The thing is that I know that this value comes from solving the characteristic equation of the homogeneous part of the equation of motion. However, I have seen another approach somewhere else. According to that procedure, the resonant frequency of this system is calculated by determining the frequency in which amplitude is maximum (zero velocity). The value obtained through that approach is similar to the frequency obtained solving the characteristic polynomial, the difference is in a 1/2 factor that multiplies the e^2. They call “displacement resonant frequency” to the frequency calculated through this procedure.

Actually, because damping is negligible, with both approaches the result would be the same (sqrt(k+ks)/m)), but now I do not understand the difference of natural frequency and displacement resonant frequency. Why are they different?

Thank you!

### RE: Natural frequency of a damped system, oscillated with a sinusoidal displacement

I have not come across the term "displacement resonant frequency".  What I gave you above is the frequency of the FREE VIBRATION response, where the ratio of damped to undamped is sqrt(1-e2).  When you look at the FORCED VIBRATION response under a harmonic loading the peak response occurs for a frequency ratio (forcing frequency divided by undamped natural frequency) of sqrt(1-2e2).  This may be what you are recalling.

### RE: Natural frequency of a damped system, oscillated with a sinusoidal displacement

(OP)
Exactly! Thank you!

Now I have my concepts more clear. I was mistaken with the "1/2" factor, I wanted to say "2". As you say, that frequency (what I called displacement resonant frequency) is the one for the forced vibration, induced by the oscillatory movement of X.

Thank you!

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