# Forced oscillation resonance frequencies [closed]

Given this forced oscillator: $$\ddot{x}+\gamma\dot{x}+\omega_0^2x=\frac{F(t)}{m}$$ Where $$F(t)$$ is: $$F(t)=\sum_{n=1}^{\infty}\frac{4F_0}{n\pi}\sin\left(\frac{2n\pi t}{T}\right) \hspace{0.4cm} \text{ for } n = 1, 3, 5, ...$$ I start with $$x_s(t)= A\sin(\omega t)+B\cos(\omega t)$$ And I solve them for A and B (see below), if I am not mistaken: $$A=\frac{F(\frac{\pi}{2\omega})}{m} \frac{\left( \omega^2-\omega^2_0\right)}{\gamma^2\omega^2+\left( \omega^2-\omega^2_0\right)} \hspace{0.2cm},\hspace{0.2cm} B=\frac{F(\frac{\pi}{2\omega})}{m} \frac{\gamma\omega}{\gamma^2\omega^2+\left( \omega^2-\omega^2_0\right)}$$

Is this solution valid?
How can I calculate the resonance frequency? Because it seems that it is not $$\omega=\omega_0$$
Is there more than one resonance frequency?

My calculations $$\dot{x}_s=\omega A\cos(\omega t)-\omega B \sin(\omega t) ~~,~~~~ \ddot{x}_s=-\omega^2 A\sin(\omega t)-\omega^2 B \cos(\omega t)$$ $$\ddot{x}+\gamma\dot{x}+\omega_0^2x=\frac{F(t)}{m}\\ \!\!\!\! =-\omega^2 A\sin(\omega t)-\omega^2 B \cos(\omega t) +\gamma\omega A\cos(\omega t)-\gamma \omega B \sin(\omega t) +\omega_0^2A\sin(\omega t)+\omega_0^2B\cos(\omega t),$$ and I solve for A and B at $$t=0$$ and $$t=\frac{\pi}{2\omega}$$.

• You can replicate the special cases in here? This is a subject covered extensively in texts. – Cosmas Zachos May 15 at 21:06
• There doesn't show a solution similar to Asin()+Bcos(). i'm asking if this could be done this way. Also, if the system is solved like this, could we find more than one resonance frequency? – Miguel NoTeimporta May 15 at 21:21
• You are mistaken in the particulars of your solution. But... after you solve the two algebraic equations right, can you guarantee that your solution works for $t=\pi/4\omega$? At other times around the circle? – Cosmas Zachos May 15 at 22:38
• – Cosmas Zachos May 16 at 16:55
• I think that part, with the Fourier series of sins, is correct. Still struggling with the resonance frequency – Miguel NoTeimporta May 16 at 17:01

Absorb the superfluous parameter m into your r.h.s. driving term, an obvious square wave with an infinity of harmonics, $$\ddot{x}+\gamma\dot{x}+\omega_0^2x= \sum_{n=1,3,5...}^{\infty}\frac{4F_0}{n\pi}\sin\left( n\omega t\right),$$ where I've defined $$\omega\equiv 2\pi/T$$, not your Ansatz parameter.
So you really only need solve the customary $$\ddot{x}+\gamma\dot{x}+\omega_0^2x= \frac{4F_0}{\pi}\sin\left( \omega t\right),$$ which you started doing with an ansatz of frequency ω, except for the full driving term, which cannot work. The steady-state solution is, then, taking the t=0 limit into consideration, $$x= -\frac{4F_0}{\pi}\left ( \sin(\omega t)\frac{\omega^2-\omega_0^2}{(\omega^2-\omega_0^2)^2+\gamma^2\omega^2}+ \cos(\omega t)\frac{\gamma\omega}{(\omega^2-\omega_0^2)^2+\gamma^2\omega^2} \right )\\ \equiv \frac{4F_0}{\pi\sqrt{(\omega^2-\omega_0^2)^2+\gamma^2\omega^2}}\bigl ( \sin(\omega t)\cos \phi + \cos(\omega t)\sin \phi \bigr ) \\ = \frac{4F_0}{\pi\sqrt{(\omega^2-\omega_0^2)^2+\gamma^2\omega^2}} \sin(\omega t + \phi) ~ , \leadsto \\ \bullet ~~~~~~~~\tan \phi ={ \gamma \omega\over \sqrt{(\omega^2-\omega_0^2)^2+\gamma^2\omega^2}. }~,$$ as customary.
The minimum of the square root denominator of the amplitude is at $$\omega=\sqrt{\omega_0^2-\gamma^2/2}~,$$ the resonant frequency of a single harmonic forcer. It is slightly below the natural frequency $$\omega_0$$ for weak friction.
Now repeat all this (in your mind) for a single forcing term of the form $$\frac{4F_0}{n\pi}\sin\left(n \omega t\right)$$, and add all these solutions to the above, by linearity, to get the general steady-state solution (that is, ignore the transient). Note each n will have a characteristic, different phase, and a characteristic resonance frequency at 1/n of the above resonant frequency. Each term will be suppressed by 1/n. Because of the different phases, all near π/4 at resonance, there won't be any mode-locking making these lower resonant frequencies invisible. In practice, you see the fundamental harmonic resonance slightly below $$\omega_0$$, but also the next odd harmonic resonance at somewhat below $$\omega_0/3$$ !