A question about damped resonant frequency

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The standard equation for the damped angular frequency of a normal damped mass-spring system is ω[itex]_{d}[/itex] = [itex]\sqrt{\frac{k}{m}-\frac{b^{2}}{4m^{2}}}[/itex]. Let p=[itex]\frac{b}{2m}[/itex], we have ω[itex]_{d}[/itex] =[itex]\sqrt{ω_{0}^{2}-p^{2}}[/itex]

Now consider that damped mass-spring system being driven by a periodic force with the driving frequency ω[itex]_{r}[/itex]. So the question is: Which value of ω[itex]_{r}[/itex] gives the biggest amplitude of the mass-spring system? (i.e. what is the damped resonant frequency of the system?)

I originally assume that for the system to resonate at its biggest amplitudes, then ω[itex]_{r}[/itex]=ω[itex]_{d}[/itex] =[itex]\sqrt{ω_{0}^{2}-p^{2}}[/itex]. However, this is not correct; and in fact the driving frequency is supposed to be ω[itex]_{r}[/itex]=[itex]\sqrt{ω_{0}^{2}-2p^{2}}[/itex]. I get this information from a video lecture from MIT open course (watch the last 1 minute of the video: )

So my question is that why is ω[itex]_{r}[/itex]=[itex]\sqrt{ω_{0}^{2}-2p^{2}}[/itex]?

Thanks in advance!!!
 
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  • #2
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Why can't you find the general solution for the damped-driven system and see where it maxes out?
 
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AlephZero
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As voko said, "do the math".

Things might get interesting when ##\sqrt{w_0^2 - p^2} > 0## but ##\sqrt{w_0^2 - 2p^2} < 0##. :devil:

I only watched the last few minutes of the video not all of it, so I'm not sure what point the lecturer was trying to make with this - but damped harmonic motion isn't a simple as you might guess, when the damping is large.
 
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damped harmonic motion isn't a simple as you might guess, when the damping is large.
When damping is large, can we even call it harmonic?
 
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AlephZero
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When damping is large, can we even call it harmonic?
Why not? The equation of motion is the same.

Certainly there is no "damped resonant frequency" for an overdanped system where ##p > w_0##, but the "maximum response" disappears when ##p > w_0/\sqrt{2}##, for a constant force input.

Physically, the reason is that your increase the forcing frequency starting below the resonant frequency, there are two opposing effects. Increasing the velocity would increase the damping and therefore the amplitude would decrease. But reducing the dynamic stiffness ##k - \omega^2 m## would tend to increase the amplitude. For low damping, the dynamic stiffness "wins". For high damping, the damping force wins.

In some engineering situations (e.g. rotating machinery) the magnitude of the force is proportional to ##\omega^2##, which leads to a different result.

But you are right that all this is not very important for many practical applications. Even for 10% damping, the differences between the undamped frequency, the damped frequency, and maximum-amplitude frequency are only of the order of 1%.
 
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I'm taking a dynamics class right now, and we use ω[itex]_{d}[/itex] = ω[itex]_{n}[/itex][itex]\sqrt{1-δ^{2}}[/itex] where δ = [itex]\frac{C}{2mω_{n}}[/itex] = damping ratio.

Ignoring the initial transient response (dependent on initial conditions) and dealing solely with the steady state response (dependent on the forcing function), the maximum response of the system can be calculated as u[itex]_{dyn}[/itex] = R[itex]_{d}[/itex]u[itex]_{static}[/itex], where R[itex]_{d}[/itex] = [itex]\frac{1}{\sqrt{Z}}[/itex], where [itex]Z=(1-β^{2})^{2}+(2βζ)^{2}[/itex] and [itex]β=\frac{ω_{force}}{ω_{n}}[/itex].

When you graph Rd with respect to beta, the response is greatest at β ≈ 1. Differentiating Z w.r.t. β

[itex]Z=1-2β^{2}+β^{4}+4ζ^{2}β^{2}[/itex]
[itex]\frac{∂Z}{∂β}=-4β+4β^{3}+8ζ^{2}β=4β^{3}+β(8ζ^{2}-4)=0[/itex]

If you solve the cubic derivative = 0, you'll get what β yields the maximum response, thus, which forcing frequency yields the largest response.
 
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