A shaft of length 2l , compressed to length 2λ , is suddenly…

A shaft of length $2l$, compressed to length $2\lambda$, is suddenly released at $t=0$ (both ends free thereafter). Find $u(x,t)$ for $-l\le x\le l$.

Step 1 · Formulate the problem

$u_{tt}=c^2u_{xx}$, free ends $u_x(\pm l,t)=0$, released from rest with initial (uniform-strain) displacement $u(x,0)=-\tfrac{l-\lambda}{l}x$, $u_t(x,0)=0$.

Step 2 · Eigenfunctions and general solution

Free-free eigenfunctions are $\cos(n\pi x/l)$, $n=0,1,2,\ldots$, so \[ u(x,t) = \frac{A_0}{2}+\sum_{n=1}^\infty A_n\cos\frac{n\pi ct}{l}\cos\frac{n\pi x}{l}. \]

Step 3 · Shift coordinates to expand the odd initial profile

Since $u(x,0)$ is odd about $x=0$, its cosine coefficients on $[-l,l]$ vanish trivially; shifting to $X=x+l\in[0,2l]$ (where $u(X,0)=(l-\lambda)(1-X/l)$ is no longer odd) and expanding in $\cos(n\pi X/2l)$ instead resolves this. Evaluating the resulting integrals gives $A_0=0$ and, for odd $n=2k-1$, \[ A_{2k-1} = \frac{8(l-\lambda)}{(2k-1)^2\pi^2},\qquad A_{2k}=0 \] (verified numerically against direct integration).

Step 4 · Shift back to $x=X-l$

Since $\cos\!\left[\tfrac{(2k-1)\pi(x+l)}{2l}\right]=(-1)^k\sin\tfrac{(2k-1)\pi x}{2l}$, the solution in the original coordinate becomes a sine series in $x$.

Verification

At $t=0$, using $\sum_{k=1}^\infty\frac{(-1)^k}{(2k-1)^2}\sin\frac{(2k-1)\pi x}{2l}=-\frac{\pi x}{8l}$ for $|x|<l$, the series reproduces $u(x,0)=-\tfrac{l-\lambda}{l}x$. \checkmark Free-end conditions: $u_x(\pm l,t)\propto\cos\tfrac{(2k-1)\pi}{2}=0$ for all $k$. \checkmark