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Let $f(x)$ be the initial triangular profile, extended by zero for $x>0$. For fixed-end reflection of a triangular pulse, the right-moving incident component is $f(x-c_{s}t)$; the reflected pulse at a fixed end has opposite sign. The full solution on $x\le 0$ isIt satisfies $y(0,t)=0$ exactly. Write $\xi =x/b\le 0$ and measure displacement in units of $a$, assuming $a,b,c_s>0$. At the first specified time the profile isAt the second time it isAt the third time the two components cancel throughout the half-string:This flat instantaneous profile does not mean the wave energy has vanished. On $-b<x<0$, its velocity at $t_3$ is $-2a{c}_s/b$, and it subsequently becomes a negative reflected pulse travelling left.

Write the initial triangular pulse as the function $h$, extended by zero to negative arguments:At $t=0$, the initial displacement and zero velocity give $f+g=h$ and $-c{f}^{\prime }+c{g}^{\prime }=0$ on the positive half-line. Fix the harmless opposite additive constants by taking both travelling waves to vanish at positive infinity. ThenThe second formula follows from the fixed-end boundary condition $f(-ct)+g(ct)=0$. Thus $f$ is one-half of the odd extension $H$ of the initial displacement, andThis is the odd-extension solution for a semi-infinite fixed string: the reflected displacement reverses sign. Evaluating the travelling profiles givesThe first profile is a positive triangle supported on $[L,3L]$, with peak $a/2$ at $2L$. The second has a negative triangle on $[0,2L]$ with minimum $-a/2$ at $L$, and a positive triangle on $[2L,4L]$ with maximum $a/2$ at $3L$.

Put the fixed end at $x=0$, with the mechanical string occupying $x\le 0$. 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 $HT(u)$, whereThe fixed-end reflection of a triangular pulse follows from the Dirichlet boundary condition $y(0,t)=0$ and the superposition principle, giving the reflected displacementThe 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 throughThe segment beside the wall is steeper because incident and inverted reflected pulses overlap there.