= Solution
<Newton's law of universal gravitation> gives the <force> $-GmM\hat{\mathbf r}/r^2$ on a test <mass> a distance $r$ from a point source. <Gauss's law for gravity> states
$$
\oint_S\mathbf g\cdot d\mathbf S=-4\pi GM_{\rm enclosed},
\qquad \nabla\cdot\mathbf g=-4\pi G\rho.
$$
The inverse-square point field has the stated flux through a surrounding sphere. Superposition extends it to distributed sources; the divergence theorem relates the integral and local forms. Symmetry is still needed to turn a flux into a pointwise field value.
For the <gravitational field of an infinite uniform cylinder>, a coaxial cylindrical Gaussian surface has no end-cap flux and constant inward field on its curved surface. The enclosed <mass> per unit length is $\pi\rho r^2$ inside and $\pi\rho R^2$ outside. Hence
$$
\boxed{g_r(r)=\begin{cases}-2\pi G\rho r,&0\le r\le R,\\-2\pi G\rho R^2/r,&r\ge R.\end{cases}}
$$
Its magnitude rises linearly from zero, reaches $2\pi G\rho R$ at the surface, then decreases as $1/r$. The field is continuous at $R$, while its radial derivative changes because the density has a sharp edge.
\Image[/past-exam-of-the-physics-course-of-the-year-4-of-the-natural-sciences-course-of-the-university-of-cambridge/2015/paper-1-cylinder-gravity.png]
{title=Gravity inside and outside an infinite uniform cylinder}
{description=Original cylindrical Gauss-law result, normalized by the surface field. The narrow drilled hole and the thin test rod are neglected as perturbations of the source field.}
Use signed distance $y$ along the diameter hole. A rod displaced by $x$ occupies $y\in[x-R,x+R]$, so its <force> is $F(x)=\lambda\int_{x-R}^{x+R}g_y(y)\,dy$. At equilibrium symmetry makes $F(0)=0$. Differentiating the moving endpoints gives
$$
F'(0)=\lambda[g_y(R)-g_y(-R)]=-4\pi G\rho\lambda R.
$$
Thus the leading restoring <force> and oscillation <frequency> are
$$
\boxed{F(x)=-4\pi G\rho\lambda R\,x+o(x),\qquad
\omega^2=\frac{4\pi G\rho\lambda R}{2R\lambda}=2\pi G\rho.}
$$
For an explicit positive-displacement check, put $C=2\pi G\rho$ and integrate the inside and outside pieces for $0<x<2R$:
$$
F(x)=-C\lambda\left[Rx-\frac{x^2}{2}+R^2\ln(1+x/R)\right].
$$
Its first-order term is $-2C\lambda Rx$, agreeing with the endpoint argument. The result is the <diameter-rod oscillation in cylindrical gravity>; it assumes a sufficiently long source cylinder and a narrow hole whose removed <mass> can be neglected.
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