I come across with this problem today. It said, when the driving force is changing, what is the peak driving frequency for maximum oscillation amplitude? It is like, asking what is the resonance frequency, right? so I though it should be the natural frequency. But no, the answer is something else.

I remember that “all forced oscillation is moving with the driving frequency”. And I don’t know that the damping will change the oscillation frequency at all! I though it is just a damping factor(term) with the oscillation!

When I search the web, many people use

\displaystyle x(t) = A \exp(-\lambda t) \cos(\omega t + \phi) ,

, where \omega is the natural frequency, as a solution, and calculated \lambda and \phi . That may be the impression that all motion has a simple frequency. But in fact, the phase factor is \omega dependent.

Here i am going to go through it again.


The motion of equation take the from

\displaystyle \frac{d^2x}{dt^2} + \omega^2 x + R \frac{dx}{dt} = A \cos(f t) ,

where R is the damping constant, and A is the magnitude of the driving force, and f is the driving frequency.

A general solution is a homogeneous solution, which is the driving term is zero, or A=0, plus a particular solution.

The homogeneous solution is

\displaystyle x(t) = C_1 e^{-\frac{R}{2}t} \cos( \omega_D t + C_2),

where

 \displaystyle \omega_D^2 = \omega^2 - \frac{R^2}{4}

and C_1, C_2 are fixed by the boundary condition. So, the damping will change the oscillation frequency. When consider the frequency is related to the energy. The damping draws the system energy to the environment, not just the oscillation amplitude, but also the frequency changed. And the frequency becomes smaller.

The particular solution is

\displaystyle x(t) = \frac{A}{f^4 + f^2 (R^2-2\omega^2) + \omega^4} ( (\omega^2 - f^2) \cos(ft) + fR \sin(ft))

We can see, when the driving force is large, the particular solution dominated, and it does not depends on any boundary condition. The oscillation frequency is f. However, look at the denominator, it can be arranged as

\displaystyle (f^2 - (\omega^2 - \frac{R^2}{2}))^2 + R^2 \omega^2 + \frac{R^4}{4}

The peak is at \omega_F = \sqrt {\omega^2 - \frac{R^2}{2}} ! not w_D, not the natural frequency!

freq_1.PNG

freq_2.PNG

omega_1.PNG


Now, lets us have two systems, P and Q.

two systems.PNG

Suppose, we already know that the damping of Q is large than P. And, by measuring, the damping frequencies are the same for P and Q too! Thus

\omega_D^2 (P) = \omega(P)^2 - R_P^2/4 = \omega(Q)^2 - R_Q^2/4 = \omega_D^2(Q)

means that the natural frequency of  Q is also larger than that of P.

R_P  < R_Q \iff \omega(P) < \omega(Q)

So, lets us set P is the group of lines on the left, which share a common natural frequency, and Q is the group of lines on the right, which share a common natural frequency too. The same color for same damping constant.

Thus, The system of P and Q, in which the damping constant of P is smaller than that of Q, we may guess something like left blue line and right orange line.

two systems_2.PNG

However, when we really fulfill the constrain, this is like this:

two systems_3.PNG

The blue curved never crossed with other curves. And it looks like figure 2. But there is difference. Let overlap two plots.

two systems_4.PNG

Suppose the curves can be crossed. That means, at some particular frequency, the system P and Q oscillating at same amplitude, but the damping of Q is larger than that of P, or the energy loss of Q is larger than that of P, given the same driving force! That is impossible!

Although the frequency of Q is larger, but the drop of amplitude is much faster, unless the damping of Q is very large. but in that case, the frequency spectrum does not have any peak at all.

p2.PNG

 

 

 

 

 

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