the ODE is:

$$\ddot x+\frac cm\,\dot x+\omega^2\,x=0$$

with $$x=2\sqrt{m\,k}\quad,\omega=\sqrt{\frac km}$$

you obtain the solution $~(x(0)=x_0~,\dot x (0)=0)~$

$$x(t)=\frac{x_0}{m}\left[{{\rm e}^{-{\frac {\sqrt {k}t}{\sqrt {m}}}}} \left( m+\sqrt {m}\sqrt { k}t \right)\right] \quad\Rightarrow\\ v(t)=-\frac{x_0}{m}\left[k{{\rm e}^{-{\frac {\sqrt {k}t}{\sqrt {m}}}}}t\right]$$

hence $~|v(t)|~$ is maximum at $~t_m=\sqrt{\frac mk}$

you can obtain the time where the velocity is maximum by solving this equation

$$\frac{d}{dt} v(t)=0$$

for t, $~\Rightarrow~t_m=t$ and

$~|v_m|=\frac{\omega\,x_0}{e}$

the ODE is:

$$\ddot x+\frac cm\,\dot x+\omega^2\,x=0$$

with $$c=2\sqrt{m\,k}\quad,\omega=\sqrt{\frac km}$$

you obtain the solution $~(x(0)=x_0~,\dot x (0)=0)~$

$$x(t)=\frac{x_0}{m}\left[{{\rm e}^{-{\frac {\sqrt {k}t}{\sqrt {m}}}}} \left( m+\sqrt {m}\sqrt { k}t \right)\right] \quad\Rightarrow\\ v(t)=-\frac{x_0}{m}\left[k{{\rm e}^{-{\frac {\sqrt {k}t}{\sqrt {m}}}}}t\right]$$

hence $~|v(t)|~$ is maximum at $~t_m=\sqrt{\frac mk}$

you can obtain the time where the velocity is maximum by solving this equation

$$\frac{d}{dt} v(t)=0$$

for t, $~\Rightarrow~t_m=t$ and

$~|v_m|=\frac{\omega\,x_0}{e}$

the ODE is:

$$\ddot x+\frac cm\,\dot x+\omega^2\,x=0$$

with $$x=2\sqrt{m\,k}\quad,\omega=\sqrt{\frac km}$$

you obtain the solution $~(x(0)=x_0~,\dot x (0)=0)~$

$$x(t)=\frac{x_0}{m}\left[{{\rm e}^{-{\frac {\sqrt {k}t}{\sqrt {m}}}}} \left( m+\sqrt {m}\sqrt { k}t \right)\right] \quad\Rightarrow\\ v(t)=-\frac{x_0}{m}\left[k{{\rm e}^{-{\frac {\sqrt {k}t}{\sqrt {m}}}}}t\right]$$

hence $~|v(t)|~$ is maximum at $~t_m=\sqrt{\frac mk}$

you can obtain the time where the velocity is maximum by solving this equation

$$\frac{d}{dt} v(t)=0$$

for t, $~\Rightarrow~t_m=t$

the ODE is:

$$\ddot x+\frac cm\,\dot x+\omega^2\,x=0$$

with $$x=2\sqrt{m\,k}\quad,\omega=\sqrt{\frac km}$$

you obtain the solution $~(x(0)=x_0~,\dot x (0)=0)~$

$$x(t)=\frac{x_0}{m}\left[{{\rm e}^{-{\frac {\sqrt {k}t}{\sqrt {m}}}}} \left( m+\sqrt {m}\sqrt { k}t \right)\right] \quad\Rightarrow\\ v(t)=-\frac{x_0}{m}\left[k{{\rm e}^{-{\frac {\sqrt {k}t}{\sqrt {m}}}}}t\right]$$

hence $~|v(t)|~$ is maximum at $~t_m=\sqrt{\frac mk}$

you can obtain the time where the velocity is maximum by solving this equation

$$\frac{d}{dt} v(t)=0$$

for t, $~\Rightarrow~t_m=t$ and

$~|v_m|=\frac{\omega\,x_0}{e}$