= Solution
“Low” and “high” compare $kT$ with the vibrational spacing $hc\widetilde\omega$, not <temperature> with an arbitrary number of kelvin. For $T\ll\theta_v$, occupation of excited levels is exponentially suppressed and
$$
\boxed{U\sim Nk\theta_v e^{-\theta_v/T}\longrightarrow0}.
$$
This vanishing result uses the ground-relative energy zero. For $T\gg\theta_v$, expand $e^{\theta_v/T}-1\sim\theta_v/T$ to obtain
$$
\boxed{U\sim NkT}.
$$
This agrees with <equipartition>: the oscillator has one quadratic kinetic and one quadratic potential term, contributing $kT/2$ each. The corresponding limiting vibrational <heat capacity> is zero at low <temperature> and $Nk$ at high <temperature>.
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