# Derivation of driven force on a string. How to prove maximum amplitud is achieved at resonant frequency?

I know if I have a driven oscillator of natural frequency $$\omega$$, applying a driven force $$F_0 \cos (\Omega t)$$ will result in a motion equation like this one (steady state/particular solution):
$$\begin{equation} x(t)=\frac{F_0}{m(\omega-\Omega)^2} \cos{\Omega t} \end{equation}$$

This shows the amplitude of oscillation increases greatly when $$\Omega \approx \omega$$.
Nevertheless, I failed to find a similar proof for a string driven by an external oscillator. Instead it's just stated "the string will oscillate with the greatest amplitude when the driven oscillator has an equal frequency to one of the normal modes given by the boundary condtion".
I was wondering if there was any explicit proof about such statement where I could actually see the amplitude of the rope as a function of the oscillator frequency $$\Omega$$. I found some proposition like "use the boundary condition of $$\psi(0,t)=A' \cos(\Omega t)$$" but the issue wasn't further developed. I actually tried this myself using the solution: $$\begin{equation} \psi(x,t)=A \sin(kx+\phi_x)\cos(\omega t +\phi_t) \end{equation}$$ but I failed.