= Solution
In <irrotational flow>, the <velocity> has a <velocity potential> locally, and globally in a simply connected flow domain. <incompressibility> then gives $\nabla\cdot\mathbf v=\nabla^2\Phi=0$. For a sphere moving at <speed> $U$ along $z$ through fluid at rest far away, the <potential of a translating sphere> is a decaying axisymmetric dipole. Its solution and its impermeability condition give
$$
\Phi=\frac{\beta\cos\theta}{r^2},\quad
\left.\partial_r\Phi\right|_a=U\cos\theta,
\qquad\boxed{\beta=-\frac12Ua^3.}
$$
These are laboratory fluid <velocities> expressed relative to a moving origin. A <velocity> field measured in the sphere-rest frame instead requires adding $-Ur\cos\theta$ to the potential.
\b[The vector identity printed in the PDF has its signs reversed.] The product rule gives the consistent identity $v^2=\nabla\cdot(\Phi\mathbf v)-\Phi\nabla\cdot\mathbf v$. With outward normal of the exterior fluid equal to $-\hat r$ at the sphere, the <divergence theorem> yields
$$
\int_{r>a}v^2\,dV=\int_{r=a}\Phi(-v_r)\,dS
=\frac{U^2a}{2}\int\cos^2\theta\,dS=\frac{2\pi a^3U^2}{3}.
$$
The integral at infinity vanishes. Hence
$$
K_f=\frac{\pi\rho_0a^3U^2}{3}=\tfrac12m_aU^2,
\qquad\boxed{m_a=\frac{2\pi\rho_0a^3}{3},\quad m_{\rm eff}=\frac{4\pi a^3}{3}(\rho+\rho_0/2).}
$$
This <added mass of a sphere> is half the displaced fluid <mass>. For a spring of unchanged stiffness, <buoyancy> only shifts the static equilibrium. Its <oscillation frequency with fluid added mass> is
$$
\boxed{\omega_s'=\omega_s\sqrt{\frac\rho{\rho+\rho_0/2}}.}
$$
For a small-angle <simple pendulum>, <buoyancy> also reduces restoring <torque> from $mg\ell\theta$ to $(m-\rho_0V)g\ell\theta$, while the inertia is $(m+m_a)\ell^2$. Thus
$$
\boxed{\omega_p'=\omega_p\sqrt{\frac{\rho-\rho_0}{\rho+\rho_0/2}}.}
$$
The downward equilibrium is stable only for $\rho>\rho_0$. The ideal results assume unbounded fluid, negligible <viscosity>, small excursions, and no important free-surface or container corrections.
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