# Damping , resonance and natural frequency?

## Main Question or Discussion Point

When a forced oscillation is damped, it is true that the frequency at which resonance occurs decreases right? so does this mean that the natural frequency has decreased too or does it only mean that resonance does not occur at natural frequency when being damped?

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Damped means with friction. Friction means additional interaction with something else. It is like dealing with a heavier oscillator - the original oscillator coupled with something else is a little bit heavier. It has its own "natural" (or proper) frequency. Of course, it is just a qualitative picture, because the damping is dissipative.

so does the natural frequency change?

Meir Achuz
Homework Helper
Gold Member
It depends on your definition of "natural frequency". By my definition, it would change.
Sometimes the term "undamped frequency" is used, and that wouldn't change.

Usually natural frequency is the frequency or frequencies that the system would oscillate at if there were no damping force or driving force (just the linear restoring force). So generally I think the natural frequency would stay the same, but the resonant frequency would be somewhere less than that in the presence of linear damping.

When a forced oscillation is damped, it is true that the frequency at which resonance occurs decreases right? so does this mean that the natural frequency has decreased too or does it only mean that resonance does not occur at natural frequency when being damped?
These simple questions show there is a lot you don't understand about resonance. And the (wrong) answers here show nobody else does either!

First. you don't say what kind of oscillator you are talking about. So for drill we'll assume the simplest case of a simple harmonic oscillator. LRC circuit, mass, spring, dashpot etc. You assume that there is a resonance here and a "natural frequency". Such may or may not be the case.

The operation of such a system depends upon the values of the various three components. What you do is look at the differential equations for the (undriven) "motion" of the various parts and solve them to give you the "natural behavior" of the system. The equation is a quadratic and you'll find that these solutions have three cases.

Two of these solutions produce solutions (roots of the "impedance") that are not "oscillations" in the normal sense at all. That would be the over-damped and critically damped cases. They produce a hump with an exponential tail (like when you jump on a car bumper). This motion (or electrical response) would be the "natural behavior" of the system under those conditions. No frequencies are to be seen unless you want to call the exponential exponents "natural frequencies".

You only get what you think of as oscillations with a frequency (damped sin waves) in the underdamped case. In this case the solutions do result in a "natural frequency". And that "natural frequency" is the frequency of the oscillatory motion of the solutions. It is NOT the frequency of the undamped system! The solutions for such a system do reduce to the sin wave of the undamped case when damping is removed, but natural frequency of motion is given by a formula where the amount of damping tends to reduce the oscillating frequency from that found with an undamped system. And the rate at which the 'natural frequency" dies out in an exponential decay also depends upon the amount of damping present.

The frequency of the undamped case is called the "undamped frequency" Duh! Or to be strictly mathematically correct it is termed the "undamped imaginary frequency". The reason being that in advanced math "generalized frequencies" are considered where even non-oscillatory exponents are considered to be "frequencies". Confused?

So to answer your question. IF you have the oscillatory under-damped case, you will have a "natural frequency" which represents the natural frequency of the freely vibrating (undriven) oscillator in response to some initial conditions. That frequency depends on the amount of damping present and the greater the damping the lower will be the "natural frequency" compared to the "natural frequency" of the undamped case. And likewise the rate of the exponential decay of the oscillations also depends upon the amount of damping with the greater damping causing the oscillations to die out sooner. (This makes sense as more energy is being dissipated hence the oscillator loses energy faster). OK?

FredGarvin
Short answer...it will probably change the frequency slightly as shown by the attached plots:

From a description of tuned isolator dampeners:
http://www.newport.com/images/web150w-EN/images/1310919.gif [Broken]

ref: http://www.newport.com/Optics-Based-ResearchThe-Need-for-Vibration-Isola/139774/1033/catalog.aspx [Broken]

The other attached image is of amplification factor vs. resonant frequency ratio for a simple forced vibration system.

In both, you can see that resonance changes very slightly for varying dampening.

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