I understand that resonance is when the force term increases the natural oscillation of the system.
In the next equation the oscillator has a natural frequency $\omega_0=\sqrt{\frac{k}{m}}$. But I don't know how to find which frequencies $\omega$ create resonance.
$$y''\left(t\right)+2\zeta y'\left(t\right)+y\left(t\right)=\sin{(\omega t+\phi)}$$
where $\zeta=\frac{c}{2\sqrt{km}}$, $c$ is the friction coefficient, $k$ the elasticity constant and $m$ the mass.
this equation has been rescaled using $y(x)=x(\omega_0t)$.
Best Answer
Although the question is not well-formulated and there are notation ambiguities, I will try to answer. The solution of the differential equation you have written consists of two parts: 1) the general solution of the homogeneous equation $y''+2\zeta y'+y=0$ and 2) a particular solution of the non-homogeneous equation. To find the first part, one should solve the characteristic equation $$\lambda^2+2\zeta\lambda+1=0\Rightarrow \lambda_{\pm}=-\zeta\pm\sqrt{\zeta^2-4}$$ and then write $y_h(t)=A\,e^{\lambda_+ t}+B\,e^{\lambda_- t}$. $A$ and $B$ are two arbitrary constants of integration.
To find a particular solution, we look for it in the form $y_{nh}(t)=\alpha\cos(\omega t+\varphi)+\beta\sin(\omega t+\varphi)$. Substituting this into the differential equation, we find two equations $$ \alpha(1-\omega^2)+2\zeta\omega\beta=0,\qquad -2\zeta\omega\alpha+\beta(1-\omega^2)=1,$$ which determine $\alpha$ and $\beta$: $$\alpha=-\frac{2\zeta\omega}{(\omega^2-1)^2+4\zeta^2\omega^2},\qquad \beta=-\frac{\omega^2-1}{(\omega^2-1)^2+4\zeta^2\omega^2}. $$ Now the general solution is $$y(t)=y_h(t)+y_{nh}(t)=A\,e^{\lambda_+ t}+B\,e^{\lambda_- t}+\alpha\cos(\omega t+\varphi)+\beta\sin(\omega t+\varphi).$$ Notice that whenever $\zeta>0$, we have $\mathrm{Re}\,\lambda_{\pm}<0$ (it doesn't matter if we have two real roots or they are complex conjugate). Therefore for large $t$ the homogeneous solution, which encodes initial conditions, becomes irrelevant. What remains is harmonic oscillations of frequency $\omega$ and amplitude $$\mathcal{A}=\sqrt{A^2+B^2}=\left[(\omega^2-1)^2+4\zeta^2\omega^2\right]^{-\frac12}$$ The resonance corresponds to the case when this amplitude is maximal. [For example, if there is no damping ($\zeta=0$) we obtain that $\omega_{res}=1$ as then the amplitude goes to infinity.] So we have to find the value of $x$ which corresponds to the minimal value of the function $$ f(x)=(x-1)^2+4\zeta^2x.$$ This gives $x_{min}=1-2\zeta^2\Rightarrow \omega_{res}=\sqrt{1-2\zeta^2}$. [Note, however, that since $x=\omega^2>0$, if $\zeta$ is sufficiently large (strong damping) then the maximal amplitude is achieved as frequency of the external force approaches $0$.]