= Solution
Put the fixed end at $x=0$, with the <mechanical string> occupying $x\le0$. Let $H$ be the incident peak <amplitude>, set $u=x/L$ and $\tau=ct/L$, and take $t=0$ when the incident front first reaches the end. The symmetric <triangular pulse> in the source has initial profile $H T(u)$, where
$$
T(s)=\max\{0,1-|2s+1|\}.
$$
The <fixed-end reflection of a triangular pulse> follows from the <Dirichlet boundary condition> $y(0,t)=0$ and the <superposition principle>, giving the reflected <displacement>
$$
\boxed{\frac{y(x,t)}H=T(u-\tau)-T(-u-\tau),\qquad u\le0.}
$$
The minus sign is the inversion on reflection. At $\tau=1/4$, the resulting profile is zero outside $[-3/4,0]$ and consists of the straight segments through
$$
\boxed{(u,y/H)=(-3/4,0),\ (-1/4,1),\ (0,0).}
$$
The segment beside the wall is steeper because incident and inverted reflected <pulses> overlap there.
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