- #1

zenterix

- 515

- 74

- Homework Statement
- Consider a harmonic oscillator driven by the function

$$f(t)=|\omega t|\tag{1}$$

Thus

- Relevant Equations
- $$\ddot{x}+\omega_n x=f(t)\tag{2}$$

The Fourier series for ##f(t)## is

$$f(t)=\frac{\pi}{2}-\frac{4}{\pi}\sum\limits_{n=1}^\infty \frac{\cos{(n\omega t)}}{n^2}\tag{3}$$

The steady-state periodic solution to the differential equation in ##x## is

$$x_p(t)=\frac{\pi}{2\omega_n^2}-\frac{4}{\pi}\sum\limits_{n=1}^\infty \frac{\cos{(n\omega t)}}{n^2(\omega_n^2-n^2\omega^2)}\tag{4}$$

The derivation of ##(4)## relied on the assumption that the natural frequency ##\omega_n## is not equal to any of the frequencies present in the Fourier series of the input signal.

Thus, ##\omega_n\neq \omega, 2\omega, 3\omega, \ldots##.

If ##\omega_n## equals any of these frequencies then resonance occurs: the response is not periodic and the amplitude blows up in time.

If ##\omega_n## does not equal any of these frequencies then, as we see in (4), we have a periodic response.

At this point, I saw the following question:

Here is the answer as it appears in these notes (answer to problem 5).

I didn't quite understand this.

Are we trying to specify an ##\omega_n## such that there are multiple different periodic ##x_p## that solve the differential equation?

Or are we after multiple different general solutions (ie, homogeneous solution plus inhomogeneous solution).

I don't think it is the latter option, since the homogeneous solution has constants so we already have infinite general solutions.

What exactly are we adding together "to get many solutions with that common period"?

$$f(t)=\frac{\pi}{2}-\frac{4}{\pi}\sum\limits_{n=1}^\infty \frac{\cos{(n\omega t)}}{n^2}\tag{3}$$

The steady-state periodic solution to the differential equation in ##x## is

$$x_p(t)=\frac{\pi}{2\omega_n^2}-\frac{4}{\pi}\sum\limits_{n=1}^\infty \frac{\cos{(n\omega t)}}{n^2(\omega_n^2-n^2\omega^2)}\tag{4}$$

The derivation of ##(4)## relied on the assumption that the natural frequency ##\omega_n## is not equal to any of the frequencies present in the Fourier series of the input signal.

Thus, ##\omega_n\neq \omega, 2\omega, 3\omega, \ldots##.

If ##\omega_n## equals any of these frequencies then resonance occurs: the response is not periodic and the amplitude blows up in time.

If ##\omega_n## does not equal any of these frequencies then, as we see in (4), we have a periodic response.

At this point, I saw the following question:

**Are there frequencies at which there is more than one periodic solution?**Here is the answer as it appears in these notes (answer to problem 5).

I didn't quite understand this.

Are we trying to specify an ##\omega_n## such that there are multiple different periodic ##x_p## that solve the differential equation?

Or are we after multiple different general solutions (ie, homogeneous solution plus inhomogeneous solution).

I don't think it is the latter option, since the homogeneous solution has constants so we already have infinite general solutions.

What exactly are we adding together "to get many solutions with that common period"?