# forced vibration question

• posted

Hello!

I have following vibration related question.

In a system under forced vibration,if the frequency of the excitation matches with the *natural frequency* of the system , the *resonance* occures and the systems starts vibrating with continuously increasing amplitudes.

My question is since energy = 1/2 K (amplitude)^2.

More amplitude means more energy.

With the same excitation force (and hence with the same input energy),how do the energy in the system go on increasing.(increase in amplitude means increase in potential energy of the spring)?

From where the system takes this energy so that the amplitude goes on increasing ?

regds, ypj

• posted

Well, it integrates the power of as many as Q cycles.

Rene

• posted

You said it yourself: "With the same excitation force (and hence with the same input energy)..."

As long as the excitation *FORCE* is present, it will provide *INPUT ENERGY* into the oscillator. Unless this energy is provided someplace to go (such as frictional heating) it will _accumulate_, *increasing* the amplitude of the vibration.

With no resistance present, if you shut off the FORCE then the amplitude will cease to *increase* and remain constant.

If there *is* resistance, the amplitude will only grow until the resistance (often a polynomial function of amplitude) drains off energy as fast as it is being pumped in by the force.

With resistance present, if you shut off the FORCE then the amplitude will gradually decrease as the energy is _depleted_ through the resistance.

Tom Davidson Richmond, VA

• posted

snipped-for-privacy@indiatimes.com (ashok) wrote in news: snipped-for-privacy@posting.google.com:

Sometimes when you can't figure a question out, it is useful to turn it on it's head and look at it backwards. If the forcing is off the natural frequency, what happens? The system vibrates at the forced frequency. What is the source of the energy to cause the system to vibrate? When you remove the forcing, does the system stop instantaneously, or does the forced vibration decay at some rate? Why?

Now, ask yourself what does natural frequency really mean?

Hint - ask yourself where the energy goes.

• posted

Only partially correct. Actually,

E = (k / 2) x^2 + (m / 2) v^2

Where v = dx / dt is the velocity. The energy, E, is constant with time, t, if the external force, F_ext = 0.

With external driving force, F_ext, The energy isn't constant with time,

dE / dt = (k x) (dx / dt) + (m v) (dv / dt)

Since, v = dx / dt, and, a = dv / dt, then

dE / dt = [(k x) + (m a)] v

but, total force, F_tot = m a = - k x + F_ext

Thus, dE / dt = v (F_ext)

This is the instantaneous rate of increase in energy, wherein both the velocity, v = v(t), and the external force F_ext = F_ext(t) are functions of time.

To get the instantaneous rate of increase in energy as a function of position, we note that

dE / dt = (dE / dx) (dx / dt) = (dE / dx) v

Thus, dE / dx = F_ext

[Old Man]
• posted

If the external force, F_ext(t), and the velocity, v(t), are in-phase (in the same direction or of the same algebraic sign), the oscillator energy increases with time. A continuous in-phase condition is possible only at the oscillator resonant frequency. However, damping will shift and broaden the oscillator's resonant condition.

[Old Man]

• posted

Work = force x distance. With proper phasing, the force is doing work on the system each cycle.

• posted

---------------- thats absolutely right yet i would like to add some physicsl understanding of it:

the misunderstanding of the orriginal poster starts with the wrong assumption that the external force is *constant*! it is not the case hasd the external force been constant it would *stop* the harmonic motion.!

the external force must be as well an harmonic force ie increase and cecreace harmonically!! now the energy is as the Rene saied is acumulated at each cycle and stored in the acted body.

we can add another question: were from the body (a body) takes its *natral energy* in order to move with its natural frequancy? in many cases to overcome friction (damping)? it is especially interesting regarding the electrons of an atom.

----------- Y.Porat

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• posted
[snip]

Wrong. Look at a spring pendulum (a vertical spring on which lower end you place a massive object, and then make it oscillate up and down). The external force of gravity is acting on this oscillations (additional to the internal force of the spring). This external force is obviously constant, but it does *not* stop the harmonic motion.

Even an engineer should know that...

If the external force is harmonic, it *can* increase the oscillations (if it has the right frequency). But it does not need to be harmonic. Being periodic is enough, if the periodic changes of the force have a Fourier component with the right frequency and a sufficient amplitude.

Err, also from outside. A spring will only oscillate if you first stretch it, i.e. add energy to it.

If there is friction, and no energy is added from outside, the oscillation will be damped, as you say. I know of no object which experiences friction, has no energy added from outside, and nevertheless goes on oscillating.

Where is friction in this case?

Bye, Bjoern

• posted

No, entirely.

Err, since energy is conserved (if F_ext = 0), (k/2) x^2 + (m/2) v^2 = 1/2 K (amplitude)^2

[snip]

Byem Bjoern

• posted

Actually, a constant force would have no effect on the oscillation IMO.

///

If you understand the interplay of potential and kinetic energy in a simple pendulum, you have the required physical insight into tuned oscillators.

Brian W

• posted

1 why are you to jump the first one on my back?? 2 a free movenent of a pendulum is damped untill stop the gravitational force os constant and yet --- it will stop and a phd should know that it is not only the vertical force of gravity but the reaction of the string *that is not constant *!!!! its direction is changing!! even a phd mathematician should know it .... so stop that patro9nising pose because you can get sometimes some lessons from an ingineer it is not a shame....

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its frequancy has to be compatible (coinciding with that one of the acted object

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ok

right yet i whanted to do the analogy with the microcosm

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thats exactly the question!! but there is 'friction' or else it will go on being at its 'excited state forever it is hard to immagine 'friction' withthe model of a rotating electron ........!!!! a chain of orbitals 9or soemething like that) is much more reasonable for that ...... have you or anyone a better suggestion??? TIA Y.Porat ps Bjoern please let other posters a chance first (with me they might not have some dull moments...... i will not let them fall sleap on their parrotong paradigmas....)

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• posted

I told you already several times that if you write nonsense, I'll correct it.

But that dampening has nothing to do with the gravitational force.

The pendulum won't stop due to the gravitational force, but due to friction.

Yes, that's entirely clear.

However, you claimed that a constant force would stop the harmonic motion. And that is simply not true.

I know it. That is entirely irrelevant for your wrong claim above that a constant force would stop the harmonic motion.

You told me nothing new here. But all what you told here is entirely irrelevant for your wrong claim above that a constant force would stop the harmonic motion.

[snip]

Err, what's your basis for thinking in the first place that there

*is* friction?

Nonsense. That excited states decay has nothing to do with friction. This can actually be explained (*quantitatively*) by vacuum oscillations. Try looking at chapter 14 of Weinberg's book on QFT, first volume.

Since no one proposes such a model, this is rather irrelevant.

If you can derive the lamb shift based on that, feel free to do so. Standard physics can.

Yes. Again, see Weinberg's book on QFT, volume one, chapter 14, for a detailed, quantitative analysis of this.

I don't hinder them in any way. If they want to post, they can.

I established *who* is the parrot here. It is not me.

Bye, Bjoern

• posted

Impeccably correct.

[Old Man]

• posted

Also the vertical weight is a sliding vector on the pendlum.

• posted

• posted

Well, that's cute.

You were really using two different meaning of "amplitute": either instantaneous, or the maximal envelope.

• posted

--------------------- i wrote a reply but it got lost so again: thank you Brian but had you known me personally you would know that i have the nasty characteristic of not being sqatisfied by just general princilpes (that are important for themselves i try as much as possible - to get into the 'bowells ' of the object ie to the more detailed situation so the detailes of the vibrating electron is what is in my mind at this point or at least a close *as much as possible model* for it. TIA Y.Porat

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• posted

--------------------- i wrote a reply but it got lost so again: thank you Brian but had you known me personally you would know that i have the nasty characteristic of not being sqatisfied by just general princilpes (that are important for themselves i try as much as possible - to get into the 'bowells ' of the object ie to the more detailed situation so the detailes of the vibrating electron is what is in my mind at this point or at least a close *as much as possible model* for it. TIA Y.Porat

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• posted

I notice that you do not bother to admit that your assertion above (that a constant force damps the oscillation) was wrong.

What makes you think that electrons vibrate?

[snip]

Bye, Bjoern

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