I had studied the damped forced vibration and had come to know that the angular speed of force which creates resonance is enter image description here

Which is counter-intuitive to me that the angular frequency should match with the damped frequency so that the force is in phase with the motion.

Therefore I am confused of the reasons behind this, I know how to differentiate a function, yet I cannot come up with a reasonable interpretation.


[...] that the angular frequency should match with the damped frequency so that the force is in phase with the motion.

That's not really true.

You're also conflating frequency and phase: two sines can have the same frequency yet different phase. I think you're a little puzzled as to why the natural oscillator frequency isn't the same as the driving force frequency. That I will address here.

The equation of motion of the driven, damped oscillator is:

$$m\ddot{x}+c\dot{x}+kx=F_0\cos(\omega t+\varphi_d)\tag{1}$$

In the underdamped case the solution is:

$$x(t)=A_he^{-\gamma t}\sin(\omega't+\varphi_h)+A\cos(\omega t-\varphi)$$

The first part of the RHS is the transient solution, the second part of the RHS is the steady state solution. $\omega'$ is the damped angular velocity (see derivation here).

It's clear that for $t\to +\infty$, the transient solution decays to zero, i.e. $A_he^{-\gamma t}\sin(\omega't+\varphi_h)=0$.

So in steady state the driving force and the motion have the same frequency: that is the frequency of the driving force ($\omega=2\pi f$)

This is the consequence of how differential equations (DE) like $(1)$ are solved. First, a particular solution is found for the homogeneous DE:


That particular solution is the transient solution.

Then a second solution is 'guessed', that is the steady state solution.

Superposition then demands the actual solution of $(1)$ is the sum (linear combination) of both solutions.

  • $\begingroup$ Thank you, but are there any explanations of why should the angular speed be equal to that? $\endgroup$
    – Derpson
    Nov 20 '16 at 17:30
  • $\begingroup$ Personally I always take the view that this is what the math tells us. Mathematical models are descriptions of the world. They often throw up solutions that are counter-intuitive but our intuition is poorly equipped to deal with nature. Go with the math and consider that to be the 'explanation', is my advice. Other approaches lead to 'I think...' type rabbit holes that don't elucidate at all. $\endgroup$
    – Gert
    Nov 20 '16 at 17:34

Your question appears to be, why is the damped natural frequency of an oscillator different from the undamped ("resonance") case? I think I can furnish you with a physical example which lets you visualize why this is true.

Imagine an oscillator in the form of a vibrating string immovably anchored at both ends. Ignore air friction. Because the ends of the string are immovable, the energy of vibration of the string cannot leave it, and it is therefore undamped. We measure its frequency of vibration and relate this to its vibrating length.

Now we replace the immovable anchor on the right hand end of the string with a little slider mechanism which allows the end of the string to move up and down a bit. We pluck the string to set it into motion. Since there's friction in the slider mechanism, we have added damping to the system, and the string's energy will be dissipated as the resonating string pulls the slider up and down.

As the string vibrates up and down, we take a snapshot of the string at that moment when the slider has been pulled upwards to its peak value of displacement. We study the photo and notice that if we extend a line towards the right through the attachment point of the string to the slider, whose slope matches that of the string to the left of its attachment point to the slider, that line will cross the zero-deflection axis of the string slightly to the right of the slider. That is, by allowing the end of the string to move up and down slightly at one end, we have effectively "lengthened" the string- which will lower the resonant frequency of the string.

In this way you can see why the damped natural frequency of the string must be slightly lower than the undamped natural frequency of the same string.


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