Rotational Barriers - Amide and Hydroxamic Acid
You can check out molecular orbitals of N-methylacetamide here. Unticking the spin-box will stop the molecule spinning. The important orbitals are $\ce{HOMO} -1$ and $\ce{HOMO} - 2$. (The actual HOMO is not part of the $\unicode[Times]{x3C0}$-system.)

The same site also features a low-level Simple Hückel Molecular Orbit Calculator. I used that to create the four images below of the $\unicode[Times]{x3C0}$-system of N-hydroxyformamide. From left to right the atoms are $\ce{O-N-C=O}$









The lowest $\unicode[Times]{x3C0}$-orbital is always bonding across the entire $\unicode[Times]{x3C0}$-system. In N-methylacetamide, the second $\unicode[Times]{x3C0}$-orbital has a nodal plain almost directly on the carbon, only slightly moved towards the carbonyl oxygen (theory tells me that, I can’t actually see that good in the model). The third $\unicode[Times]{x3C0}$-orbital is empty and thus does not participate in bonding. This means that each bond ($\ce{N-C}$ and $\ce{C=O}$ can be regarded as something between a single and a double bond — in accordance with the Lewis resonance structures. There is considerable overlap to be lost on rotating the system around $\ce{N-C}$.

In the hydroxamic acid, the second and third $\unicode[Times]{x3C0}$-orbitals (the occupied ones) only have very little electron density in the $\ce{N-C}$ bond. In fact, the lowest MO is largest for the $\ce{O-N}$ bond, the second is (weakly) antibonding concerning $\ce{O-N}$ and has the largest contributions on $\ce{C=O}$. The third is again antibonding along $\ce{O-N}$. Altogether this means that most of the electron density is between $\ce{O-N}$ and between $\ce{C=O}$ — the $\ce{N-C}$ bond only carries very little density and is thus much better described by a single bond. Much less resonance is lost when rotating, so the rotational barrier is lower.