# Resonance of a damped harmonic oscillator under forced oscillations

Suppose we have a damped harmonic oscillator and we also apply an external force such that our system oscillates in steady state. If the frequency of my force matches the natural frequency of my oscillator, I will have resonance. However, I've also read the description of resonance as the unidirectional flow of energy from the external force to the oscillator. Can anyone explain what this exactly means and how?

Also, I've read other posts here, where they use fourier transform of force while deriving x. I have no idea why Fourier transform is used here, and how does it make the calculations any simpler. Please illuminate me on this!!

the unidirectional flow of energy from the external force to the oscillator

means that the external force is giving the oscillator energy to balance the energy lost in the dampening. By knowing the general solution $$x(t) = A_1 e^{-\gamma t} \cos(\omega_1 t + \phi_1) + A_2 \cos(\omega_2 t + \phi_2)$$ In stationary condition it becomes ($$t \rightarrow \infty$$) $$x(t) = A_2 \cos(\omega t + \phi)$$ we can now differentiate and get $$\dot x (t) -\omega A_2 \sin(\omega t + \phi_2)$$ Energy is the sum of kinetic and potential energy: $$E(t) \frac{1}{2} m A_2^2 \omega_0^2 \cos^2(\omega t + \phi_2) + \frac{1}{2}mA_2^2 \omega^2 sin^2(\omega t + \phi_2)$$

In resonance conditions ($$\omega = \omega_0$$) we get $$E (t) = \frac{1}{2} m A_2 \omega_0^2$$ Energy does not depend on time; this means that all the energy lost because of the dampening is immediately brought back by the external force. I guess this is what your text book means with 'unidirectional flow of energy'.

Hope this helps.

If the external force frequency is different from the resonant frequency, the force does positive work for part of each oscillation cycle and negative work for the other part.

To see what this means physically, suppose the force frequency applied to a spring and mass system is very low. For half the cycle, the force does work compressing the spring. For the other half, the spring does work as it is released.

At resonance, the force is 90 degrees out of phase with the displacement, and the force does positive work for the complete cycle.

I have no idea what you mean by "using fourier transform of force while deriving $$x$$." For example do you really mean "derive" or are you making the common mistake of writing "derive" when you mean "differentiate"? (And what does $$x$$ represent anyway?)

• There was a mistake. I did mean deriving. But, in this context: – Ruchi Feb 12 at 11:30
• Consider a generic damped harmonic oscillator. There are three forces, the restoring force $F_\text{restoring} = - k x(t)$, the friction force $F_\text{friction} = - \mu \dot{x}(t)$, and the driving force $F_\text{drive}(t)$. Newton's law says $F(t) = m \ddot{x}(t)$ which gives $$-k x(t) - \mu \dot{x}(t) + F_\text{drive}(t) = m \ddot{x}(t) \, .$$ – Ruchi Feb 12 at 11:31
• Dividing through by $m$ and defining $\phi(t) \equiv x(t)/m$, $\omega_0^2 \equiv k/m$, $2 \beta \equiv \mu/m$, and $J(t) \equiv F_\text{drive}(t)/m$, we get $$\ddot{\phi}(t) + 2\beta \dot{\phi}(t) + \omega_0^2 \phi(t) = J(t) \, .$$ – Ruchi Feb 12 at 11:32
• Writing $\phi(t)$ as a Fourier transform $$\phi(t) = \int_{-\infty}^\infty \frac{d\omega}{2\pi} \, \tilde{\phi}(\omega) e^{i \omega t}$$ and plugging into the equation of motion, we find $$\left( - \omega^2 + i 2 \beta \omega + \omega_0^2 \right) = \tilde{J}(\omega)$$ which can be rewritten as $$\tilde{\phi}(\omega) = - \frac{\tilde{J}(\omega)}{\left( \omega^2 - i 2 \beta \omega - \omega_0^2 \right)} \, .$$ – Ruchi Feb 12 at 11:32
• Let's take the case where the drive is a cosine, i.e. $J(t) = A \cos(\Omega t)$. In this case $\tilde{J}(\omega) = (1/2)\left(\delta(\omega - \Omega) + \delta(\omega + \Omega) \right)$ so if you work it all out you find $$\phi(t) = \Re \left[ - \frac{A e^{i \Omega t}}{\Omega^2 - i 2 \beta \Omega - \omega_0^2} \right] \, .$$ It's easy to check that $\phi(t)$ has the largest amplitude when $\Omega = \omega_0 \sqrt{1 - 2 (\beta / \omega_0)^2}$, which decreases as $\beta$ increases. – Ruchi Feb 12 at 11:34