A shaft fixed at x=0 carries a tensile force P at the free…
A shaft fixed at $x=0$ carries a tensile force $P$ at the free end $x=l$, suddenly removed at $t=0$. Find $u(x,t)$ for $t>0$.
Explanation of Method
The pre-release static displacement is linear; for $t>0$ this becomes the initial condition for free vibration of a fixed-free bar, whose eigenfunctions are odd-quarter-wave sines.
Step 1 · Static initial displacement
Before release, $EAu''=0$ with $u(0)=0$, $EAu'(l)=P$ gives $u_{\rm st}(x)=Px/(EA)$.
Step 2 · Fixed-free eigenfunctions
With $u(0,t)=0$, $u_x(l,t)=0$: $X=A\sin\mu x$ and $\cos\mu l=0$ give \[ \mu_n=\frac{(2n-1)\pi}{2l},\qquad X_n(x)=\sin\frac{(2n-1)\pi x}{2l}. \]
Step 3 · General solution and coefficients
Since $u_t(x,0)=0$: \[ u(x,t)=\sum_{n=1}^\infty A_n\cos\frac{(2n-1)\pi ct}{2l}\sin\frac{(2n-1)\pi x}{2l}, \] with $A_n$ fixed by $u(x,0)=Px/(EA)$; evaluating the integral ($\int x\sin\mu x\,dx=\sin\mu x/\mu^2-x\cos\mu x/\mu$) gives \[ A_n = \frac{8Pl(-1)^{n+1}}{EA(2n-1)^2\pi^2} \] (verified numerically against direct integration).
Verification
Using $\sum_{n=1}^\infty\frac{(-1)^{n+1}}{(2n-1)^2}\sin\frac{(2n-1)\pi x}{2l}=\frac{\pi x}{8l}$ on $[0,l]$, the series reproduces $u(x,0)=Px/(EA)$. \checkmark Boundary conditions $u(0,t)=0$ and $u_x(l,t)\propto\cos\tfrac{(2n-1)\pi}{2}=0$ hold for all $n$. \checkmark Only odd harmonics appear, as expected for a fixed-free bar.
Result
The solution is: