A thin, uniform rod has mass $m$ and length $L$. Let the acceleration due to gravity be $g$. Let the rotational inertia of the rod about its center be $md^2$.

The rod is suspended from a distance $kd$ from the center, and undergoes small oscillations with an angular frequency $\beta \sqrt{\frac{g}{d}}$.

Find an expression for $\beta$ in terms of $k$.

$ \textbf{(A)}\ 1+k^2$

$ \textbf{(B)}\ \sqrt{1+k^2}$

$ \textbf{(C)}\ \sqrt{\frac{k}{1+k}}$

$ \textbf{(D)}\ \sqrt{\frac{k^2}{1+k}}$

$ \textbf{(E)}\ \text{none of the above}$
1. **Determine the rotational inertia of the rod about the suspension point:**

   The rotational inertia \( I \) of the rod about its center is given as \( I_{cm} = md^2 \). When the rod is suspended from a point \( kd \) from the center, we use the parallel axis theorem to find the new rotational inertia:
   \[
   I = I_{cm} + m(kd)^2 = md^2 + mk^2d^2 = m(1 + k^2)d^2
   \]

2. **Set up the equation of motion for small oscillations:**

   When the rod is at an angle \( \phi \) with the vertical, the torque \( \tau \) due to gravity is:
   \[
   \tau = -mg(kd) \sin \phi
   \]
   For small angles, \( \sin \phi \approx \phi \), so:
   \[
   \tau \approx -mg(kd) \phi
   \]

3. **Apply Newton's second law for rotation:**

   Newton's second law for rotation states that the torque is equal to the moment of inertia times the angular acceleration:
   \[
   \tau = I \ddot{\phi}
   \]
   Substituting the expressions for \( \tau \) and \( I \):
   \[
   -mg(kd) \phi = m(1 + k^2)d^2 \ddot{\phi}
   \]

4. **Simplify the equation:**

   Divide both sides by \( m(1 + k^2)d^2 \):
   \[
   -\frac{kg}{(1 + k^2)d} \phi = \ddot{\phi}
   \]

5. **Identify the angular frequency \( \omega \):**

   The equation \( \ddot{\phi} + \omega^2 \phi = 0 \) describes simple harmonic motion, where:
   \[
   \omega^2 = \frac{kg}{(1 + k^2)d}
   \]
   Therefore:
   \[
   \omega = \sqrt{\frac{kg}{(1 + k^2)d}}
   \]

6. **Relate \( \omega \) to the given form \( \beta \sqrt{\frac{g}{d}} \):**

   Given that \( \omega = \beta \sqrt{\frac{g}{d}} \), we equate the expressions:
   \[
   \beta \sqrt{\frac{g}{d}} = \sqrt{\frac{kg}{(1 + k^2)d}}
   \]

7. **Solve for \( \beta \):**

   Square both sides to eliminate the square roots:
   \[
   \beta^2 \frac{g}{d} = \frac{kg}{(1 + k^2)d}
   \]
   Cancel \( \frac{g}{d} \) from both sides:
   \[
   \beta^2 = \frac{k}{1 + k^2}
   \]
   Take the square root of both sides:
   \[
   \beta = \sqrt{\frac{k}{1 + k^2}}
   \]

Thus, the correct expression for \( \beta \) is:
\[
\boxed{\sqrt{\frac{k}{1 + k^2}}}
\]