= Solution
Use the <Euler-Bernoulli beam equation> for a uniform slender <elastic beam>, small deflection and small slope, neglecting axial tension, shear deformation, rotational inertia of a cross-section and <damping>. Let $E$ be <Young's modulus>, $I=\int_A z^2dA$ the <second moment of area> about the <neutral axis>, and $\rho$ the <mass> per unit <length>. If <density> instead denotes <mass> per <volume>, replace this $\rho$ by $\rho_{\rm bulk}A$. The <bending stiffness> is $EI$; the small-slope <curvature> is $y_{xx}$. The bending and <kinetic energies> are
$$
U=\frac{EI}{2}\int_0^L y_{xx}^2dx,\qquad
T=\frac\rho2\int_0^L y_t^2dx.
$$
Twice integrating the variation of $U$ by parts gives the restoring load $-EIy_{xxxx}$ per unit <length>. <Newton's second law> therefore gives
$$
\boxed{\rho y_{tt}+EIy_{xxxx}=0.}
$$
At a pivot, <displacement> is fixed but slope is free; zero end <bending moment> gives $EIy_{xx}=0$. This also follows from the natural boundary term $EIy_{xx}\delta y_x$ in the variation.
With <separation of variables>, $y=XT$ gives
$$
\frac{T''}{T}=-\frac{EI}{\rho}\frac{X''''}{X}=-\omega^2.
$$
Choose the time normalization $T(0)=1$. Initial rest removes the <sine> time term, so $T(t)=\cos\omega t$. The spatial characteristic roots are $\pm ik,\pm k$, giving
$$
X=A\sin kx+B\cos kx+C\sinh kx+D\cosh kx,
\qquad
\boxed{\omega=k^2\sqrt{EI/\rho}.}
$$
The two conditions at $x=0$ give $B+D=0$ and $-B+D=0$, hence $B=D=0$. At $L$, addition and subtraction of $X(L)=X''(L)=0$ give $C\sinh kL=0$ and $A\sin kL=0$. Nontrivial positive-frequency modes therefore have $C=0$ and $k_n=n\pi/L$. Thus the <simply supported beam normal modes> are
$$
\boxed{X_n(x)=\sin\frac{n\pi x}{L},\qquad
\omega_n=\frac{n^2\pi^2}{L^2}\sqrt{\frac{EI}{\rho}},\quad n=1,2,\ldots.}
$$
The zero-frequency cubic solution also vanishes under these four endpoint conditions, so $n=0$ is not an additional mode.
For the specified release, small slope requires $|a|L\ll1$. Expand the initial deflection in the <Fourier sine series>. Its coefficients are
$$
A_n=\frac{2a}{L}\int_0^L x(x-L)\sin\frac{n\pi x}{L}\,dx
=-\frac{4aL^2[1-(-1)^n]}{n^3\pi^3}.
$$
Writing $\omega_1=\pi^2\sqrt{EI/\rho}/L^2$, the resulting motion is
$$
\boxed{y(x,t)=-\frac{8aL^2}{\pi^3}\sum_{j=0}^\infty
\frac{\sin[(2j+1)\pi x/L]}{(2j+1)^3}
\cos[(2j+1)^2\omega_1t].}
$$
Only odd <normal modes> occur because the initial profile is symmetric about the midpoint, and their <amplitudes> decrease as $n^{-3}$. The <frequencies> grow as $n^2$, so the motion is dispersive and is not a single shape multiplied by one <cosine>. Nevertheless the ideal undamped solution is exactly periodic with period $2\pi/\omega_1$. At $t=\pi/(2\omega_1)$ all odd-mode <cosines> vanish, so the bar is instantaneously straight; at $t=\pi/\omega_1$ it has the opposite initial deflection. The first mode dominates <displacement>, with third-mode <amplitude> only $1/27$ of the first.
There is <initial curvature incompatibility at a moment-free beam end>: the prescribed <parabola> has $y_{xx}(x,0)=2a$, whereas a moment-free pivot requires zero <curvature> at an endpoint. Therefore the data do not define a classical solution with <curvature> continuous through the two end-time corners. They do define a finite-energy <weak solution> after release of the <forces> or end moments that prepared the initial shape. The <sine> expansion represents that solution: its <displacement> reproduces the initial <parabola>, each mode satisfies the released endpoint conditions, and $\sum_n n^4|A_n|^2<\infty$ gives finite bending <energy>. One must not differentiate the initial series twice at its endpoints and assign the interior <parabola>'s <curvature> there. This qualification leaves the modal <displacement> prediction intact.
\Image[/past-exam-of-the-physics-b-course-of-the-university-of-cambridge/2012/paper-1-beam-release.png]
{title=Original modal evolution after releasing a parabolic <elastic beam> deflection}
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