The frequency of small oscillations of thin uniform vertical rod of...

Let the rod be displaced by a small angle $\theta$ from the vertical.

The horizontal displacement of the lower end of the rod is $x = l\theta$.

The restoring force exerted by the two springs is $F = (k_1 + k_2)x = (k_1 + k_2)l\theta$.

The restoring torque due to the springs about the hinge O is:
$\tau_s = -F \cdot l = -(k_1 + k_2)l^2\theta$

The center of mass of the rod is at a distance $\dfrac{l}{2}$ from the hinge O. The restoring torque due to gravity about O is:
$\tau_g = -mg \left(\dfrac{l}{2}\right) \sin\theta \approx -\dfrac{mgl}{2}\theta$

The total restoring torque on the rod is:
$\tau = \tau_s + \tau_g = -\left[ (k_1 + k_2)l^2 + \dfrac{mgl}{2} \right]\theta$

The moment of inertia of the uniform rod about the hinge O is $I = \dfrac{ml^2}{3}$.

Using the equation of motion $\tau = I\alpha$:
$\dfrac{ml^2}{3}\alpha = -\left[ (k_1 + k_2)l^2 + \dfrac{mgl}{2} \right]\theta$

$\alpha = -\left[ \dfrac{3(k_1 + k_2)}{m} + \dfrac{3g}{2l} \right]\theta$

Comparing this with the standard equation for simple harmonic motion $\alpha = -\omega^2\theta$, the angular frequency is:
$\omega = \sqrt{\dfrac{3(k_1 + k_2)}{m} + \dfrac{3g}{2l}}$

The frequency of oscillation is:
$f = \dfrac{\omega}{2\pi} = \dfrac{1}{2\pi}\sqrt{\dfrac{3(k_1 + k_2)}{m} + \dfrac{3g}{2l}}$

Answer: $\dfrac{1}{2\pi}\sqrt{\dfrac{3(k_1 + k_2)}{m} + \dfrac{3g}{2l}}$