= Solution
Use <separation of variables> $u=X(x)T(t)$. The fixed-end <Dirichlet boundary conditions> select $X_n=\sin(nx)$, $n\geq1$, and each mode satisfies $T_n''+2hT_n'+n^2T_n=0$. In the physical resistance range $0\leq h<1$, all <normal modes> are underdamped, with $\omega_n=\sqrt{n^2-h^2}>0$. The modal factor normalized by $T_n(0)=1$, $T_n'(0)=0$ is
$$
T_n(t)=e^{-ht}\left[\cos(\omega_nt)+\frac h{\omega_n}\sin(\omega_nt)\right].
$$
Consequently the <linearly damped string> has solution
$$
\boxed{u(x,t)=\sum_{n=1}^\infty b_n\sin(nx)e^{-ht}\left[\cos(\omega_nt)+\frac h{\omega_n}\sin(\omega_nt)\right],\quad b_n=\frac2\pi\int_0^\pi f(x)\sin(nx)dx.}
$$
The coefficients are those of the initial <Fourier sine series>. Differentiating each normalized modal factor at zero confirms that the string starts from rest. Suitable initial data give convergence in the string's energy space; the piecewise linear profile here need not give a twice differentiable solution along every characteristic from its corner.
For the specified tent profile, integrating its two linear pieces by parts gives
$$
\int_0^\pi f(x)\sin(nx)dx=\frac{2}{n^2}\sin\frac{n\pi}{2}.
$$
For example the first piece gives $-\pi\cos(n\pi/2)/(2n)+\sin(n\pi/2)/n^2$, and the second gives the opposite cosine term and the same sine term. Hence
$$
\boxed{b_n=\frac4{\pi n^2}\sin\frac{n\pi}{2},\qquad b_{2m}=0,\quad b_{2m+1}=\frac{4(-1)^m}{\pi(2m+1)^2}.}
$$
The zero even coefficients also follow from reflection symmetry about the midpoint. This is the <Fourier sine series of a symmetric tent>. If $h<1$ were read without the physical assumption $h\geq0$, the same modal solution extends by replacing imaginary-frequency sine and cosine by their hyperbolic counterparts, and taking $\sin(\omega t)/\omega=t$ when $\omega=0$.
\Image[/past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2009/paper-1-damped-tent-string.png]
{title=Evolution of a tent-shaped string released from rest with linear damping}
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