Step 1: Write down the normalized ground-state wavefunction.
The ground-state wavefunction of the 1D harmonic oscillator is a Gaussian:
\[ \psi_0(x) = \left(\frac{\alpha}{\pi}\right)^{1/4}e^{-\alpha x^2/2}, \qquad \alpha=\frac{2\pi m\nu}{h} \]
where $m$ is the oscillator mass.
Step 2: Compute the average potential energy directly.
The potential is $V(x)=\frac{1}{2}m\omega^2x^2$ with $\omega=2\pi\nu$. Using $\langle x^2\rangle=\frac{\hbar}{2m\omega}$ for this normalized $\psi_0$:
\[ \langle V\rangle = \frac{1}{2}m\omega^2\cdot\frac{\hbar}{2m\omega} = \frac{\hbar\omega}{4} = \frac{h\nu}{4} \]
since $\hbar\omega=h\nu$.
Step 3: Get the average kinetic energy from the total minus potential.
The ground-state total energy is $E_0=\frac{1}{2}h\nu$ regardless of how it splits. Since $\langle T\rangle+\langle V\rangle=E_0$:
\[ \langle T\rangle = E_0-\langle V\rangle = \frac{h\nu}{2}-\frac{h\nu}{4} = \frac{h\nu}{4} \]
This matches a direct momentum-space computation, $\langle p^2\rangle/2m$ for the same Gaussian also gives $\hbar\omega/4$, confirming the two averages are equal by explicit integration.
Step 4: Match to the options.
Both $\langle T\rangle$ and $\langle V\rangle$ equal $h\nu/4$, so the two averages are equal and each is a quarter of $h\nu$, not half.
Final Answer:
Option (C): $h\nu/4$ and $h\nu/4$, respectively.
\[\boxed{\langle T\rangle=\langle V\rangle=\frac{h\nu}{4}}\]