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At the steady state, the average stored energy is given by $$E = \dfrac{1}{2} m\left\langle (x')^2\right\rangle + \dfrac{1}{2} m{\omega_0}^2 \left\langle x^2 \right\rangle$$

They become equal when the driving frequency $\omega = \omega_0$.

Here is the qualitative explanation given by Frank S. Crawford in his book:

If $\omega$ is large compared with $\omega_0$, the velocity of $m$ gets reversed before it has a chance to acquire large displacement & hence it can acquire a large potential energy.

On the other hand, if $\omega$ is small compared with $\omega_0$, the velocity never gets very large, & then the time_average potential energy dominates.

I'm having some problem with the explanation. Suppose if I give initially a certain velocity to the oscillator, the restoring force works against it & if the spring-constant is more, it only takes less time to bring the oscillator to rest converting the kinetic energy completely to potential energy in a lesser time than the time required if the spring-constant were smaller than $\omega$ . So, I really couldn't understand the first explanation. Same is with the second also. I am failing to visualise what he is saying. Please help me explain his words

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  • $\begingroup$ I don't understand your problem. Given that the two are equal at resonance, the explanation given is very clear - for $\omega \gg \omega_0$, the average kinetic energy is far higher than at resonance (because $\omega$ is faster oscillation than $\omega_0$), but the average potential lower, and vice versa for $\omega \ll \omega_0$. $\endgroup$
    – ACuriousMind
    Commented Jun 11, 2015 at 10:19
  • $\begingroup$ @ACuriousMind: If it has higher kinetic energy, doesn't it get converted to higher potential energy? This is my problem. $\endgroup$
    – user36790
    Commented Jun 11, 2015 at 10:23
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    $\begingroup$ The driven harmonic oscillator is a dissipative system - the driving mechanism is constantly doing work on the oscillator, the total energy is not conserved at all times (only on average away from the resonant case), so your argument for why the average potential and kinetic energies should be the same fails (because it is based on the sum of kinetic and potential energy being the same always). $\endgroup$
    – ACuriousMind
    Commented Jun 11, 2015 at 10:35

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ACuriousMind is right. Assuming the system contains a dissipative component, for any fixed driving frequency, when the rate of energy input from the driving source equals the rate at which energy is lost from any dissipative elements in the system, a dynamic equilibrium is reached in the system. At dynamic equilibrium the sum of the average potential and average kinetic energy are a constant.

What Crawford is trying to illustrate is what particular form of energy is dominant below and above the natural (resonant) frequency of the system. Below the resonant frequency kinetic energy is dominant and above the resonant frequency potential energy is dominant. And at the resonant frequency each form is equal.

Note that prefaced "For any fixed driving frequency ..." above. The total energy at dynamic equilibrium is a constant, but only at a fixed frequency. The total energy is a maximum at the resonant frequency. At the resonant frequency, the system has its maximum capacity for energy.

Now if there is no energy dissipation or the rate of dissipation is less than the rate of energy input - there is danger of the system 'blowing up'. But in real physical systems there is always nonlinearity. The nonlinearity may provide a safe release for energy once some limiting factor is reached, or in other cases (like the Tacoma Narrows Bridge) reach a point of disaster.

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  • $\begingroup$ Let me sir, make my problem a bit clear. As is evident from Mr. Crawford's words, the driving force maintains the kinetic energy & the potential energy is maintained by spring-constant. When the frequency is lower than natural frequency, the kinetic energy is larger. Doesn't this larger kinetic energy get converted to potential energy when the oscillator reaches the turning-point?? If you give initially certain velocity to a mass attached to a spring, the whole kinetic energy gets converted to potential energy at the turning point. .... $\endgroup$
    – user36790
    Commented Jun 11, 2015 at 15:54
  • $\begingroup$ .... ,isn't it? At least I have read this. Here the kinetic energy is more & potential energy is less. I'm just not understanding this. As the spring expands, the velocity of the mass & thus the greater kinetic energy decreases to zero, increasing the potential energy. If the kinetic energy is larger, that means it is not completely converting to potential energy, is it? I request you sir, to please describe the scenario, please. $\endgroup$
    – user36790
    Commented Jun 11, 2015 at 16:31
  • $\begingroup$ Sorry, a typo, above: 'larger' in place of 'lower'. $\endgroup$
    – user36790
    Commented Jun 11, 2015 at 17:11
  • $\begingroup$ @user36790 kinetic energy is defined when there is motion (a velocity or better put momentum). Potential energy, when there is no velocity but rather energy being trapped in a field. When you 'lump' ideal elements in a spring mass oscillator the spring has no mass and the mass has no field. So you can clearly say where the energy is going - in the ideal model. But in reality all components have mass and elasticity so its a little less clear where the energy is going - but ideally can be accounted for. $\endgroup$
    – docscience
    Commented Jun 11, 2015 at 18:45
  • $\begingroup$ @user36790 Also the driving force does not maintain kinetic energy. It's usually considered as just a plain energy input. Within the system being drive, kinetic energy is maintained when the mass is in motion and potential energy is maintained when the spring is deflected. $\endgroup$
    – docscience
    Commented Jun 11, 2015 at 18:49

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