Do protons or neutrons oscillate? (inside nucleus in atom) Do they oscillate relative to each other? What is the frequency? What is the amplitude? I would think they oscillate since electrons move all over the place at high speeds and there is attractive Coulomb force between electrons and protons.
 A: Protons and neutrons in nuclei interact mainly via strong and weak forces, which is why the excitation energies of nuclei (i.e., the frequencies with which protons and neutrons "oscillate") are usually much bigger than those associated with electrons, which interact with protons via electromagnetic interaction (that is the Coulomb forces, strong, weak and Coulomb interactions are different kinds of fundamental interactions).
Thus, characteristically, the atomic spectra associated with movement of electrons lie in infrared, visible and sometimes X-ray region, whereas those of nuclei are in gamma-ray region (see also Do nuclei emit photons?).
A: Given that nucleon is bound to nuclei radius,- this per Heisenberg uncertainty, gives us uncertainty in nucleon speed, which is :
$$ \Delta v = \frac {ℏ}{2~m_n R_0  A^{1/3}} $$
Where $m_n$ is nucleon mass, $R_0 = 1.2×10^{−15}~m$ and $A$ is nucleon amount in specific atom.
If we would calculate speed uncertainty for neutron in Iron nucleus, this equation would give that such neutron would jiggle around with $ \approx 7000 \text {km/s} $ uncertainty in speed, which is quite high.
To get representation of neutron oscillation frequency, imagine that this speed is tangential speed of neutron revolving around nuclei, which is not quite a good model, but gives fast estimates. In this case you can use angular frequency and tangential speed relationship :
$$ v_{_\perp} = \omega~r $$
to extrapolate neutron oscillation frequency order in Iron atom, which would be on the order $$ f \approx 10^{21}~\text{Hz} $$.
Which is amazingly huge. I can't say if this is true, but certainly nucleons in a nuclei aren't something hammered to exact position like raisins in a bread. They are more describable by a quantum wavefunction, which probably could be calculated to complete nuclei as well.
