Transversal string with free ends question

We are given a string of length $L$ tied to massless rings at both ends so as they can move freely in the transversal direction. The linear density of the string is $ρ$ and the tension at rest (where $y=0$ for the whole string) is $T$.

I found that the normal modes of the string are: $$y_n(x,t)=A\cos(k_nx)\cos(\omega_nt+\phi_n)$$ where $k_n={\pi n \over L}$ and $\omega_n=k_n \sqrt{T \over \rho}$

When the string is at rest, a transversal velocity $v_0$ is momentarily applied at the $x=L$ end of the string, so all points of the string have a distribution of linearly reducing speeds, with $x=0$ having a speed of 0. I must express the motion of the string as a superimposition of all the normal modes.

What I was thinking is: If the speed at $x=0$ is $0$, this means that $${\partial y\over \partial t }\bigg|_{x~=~0} =0 \Leftrightarrow -A\omega \sin(\omega t+\phi )=0 \Leftrightarrow \omega t+\phi=n \pi \Leftrightarrow \phi=n\pi-\omega t$$ This gives $y_n(x,t)=A\cos(kx)\cos(n\pi)$, which is not a function of $t$.
Am I wrong in some way?

You know that $$y(x,t)=\sum\limits_{n=1}^{\infty} y_n(x,t)=\sum\limits_{n=1}^{\infty} A_n\cos(k_nx)\cos(\omega_nt+\phi_n)$$

and

$$\dot y(x,t)=\sum\limits_{n=1}^{\infty} \dot y_n(x,t)=\sum\limits_{n=1}^{\infty} -A_n\omega_n\cos(k_nx)\sin(\omega_nt+\phi_n)$$

where $k_n={\pi n \over L}$ and $\omega_n=k_n \sqrt{T \over \rho}$.

Given $y(x,0)=0$ and $\dot y(x,0)= v_o\dfrac xL$ as the two initial conditions you need to solve for $A_n$ and $\phi_n$ using Fourier analysis as described here and in many other texts.

If $y_n(x,t)=A\cos(k_nx)\cos(\omega_nt+\phi_n)$ and you want $y_n$ to be zero when $x = 0$ the only that's going to happen is if $A=0$, no?

The velocity profile you describe with one end moving while the other is stationary does not go with the the equation for the normal modes. In that equation the velocity at $x=0$ is always equal in magnitude to the velocity at $x=L$.