= Solution
For each sine <normal mode>, the temporal <characteristic equation> is $s^2+2\lambda s+\alpha_n^2=0$, with roots $s=-\lambda\pm\sqrt{\lambda^2-\alpha_n^2}$. Since $\lambda L/\pi=M+1/2$, the first $M$ modes are <overdamped> and all higher modes are <underdamped>. No integer $n$ gives <critical damping>.
The zero initial displacement excludes the hyperbolic cosine and cosine terms. Hence
$$
\begin{aligned}
u(x,t)=e^{-\lambda t}\bigg[&\sum_{n=1}^{M}A_n\sin(\alpha_nx)\sinh(\Omega_nt)\\
&+\sum_{n=M+1}^{\infty}A_n\sin(\alpha_nx)\sin(\omega_nt)\bigg],
\end{aligned}
$$
where
$$
\boxed{\begin{aligned}
\alpha_n&=\frac{n\pi}{L},\\
\Omega_n&=\frac\pi L\sqrt{\left(M+\frac12\right)^2-n^2}\quad(1\leq n\leq M),\\
\omega_n&=\frac\pi L\sqrt{n^2-\left(M+\frac12\right)^2}\quad(n\geq M+1).
\end{aligned}}
$$
To impose the initial <velocity>, let $f_n=\frac2L\int_0^L f(x)\sin(\alpha_nx)\,dx$ be its <Fourier sine series> coefficients. Differentiating the expansion at $t=0$ gives $A_n\Omega_n=f_n$ for the <overdamped> modes and $A_n\omega_n=f_n$ for the <underdamped> modes. These determine all coefficients. Despite the hyperbolic sine, the first modes decay: $\Omega_n<\lambda$, so both temporal exponential rates are negative.
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