Pivoting planks.

Relevant wiki: simple harmonic oscillator

Plank kinetic energy:

$E_{plank} = \frac{1}{2} \big(\frac{m L^2}{3}\big) \dot{\theta}^2$

Mass kinetic energy:

$E_{mass} = \frac{1}{2} M L^2 \dot{\theta}^2$

Plank gravitational potential energy relative to rest position:

$U_{plank} = \frac{1}{2} \, mg \, L \sin\theta$

Mass gravitational potential energy relative to rest position:

$U_{mass} = Mg \, L \sin\theta$

Spring potential energy relative to rest position:

$U_{spring} = \frac{1}{2} k L^2 \sin^2\theta$

Forming the Lagrangian and using the small angle approximation ( $\sin\theta \approx \theta)$ :

$L = E_{total} - U_{total} = \frac{1}{2} \big(\frac{m L^2}{3}\big) \dot{\theta}^2 + \frac{1}{2} M L^2 \dot{\theta}^2 - \frac{1}{2} \, mg \, L \theta - Mg \, L \theta - \frac{1}{2} k L^2 \theta^2$

Equation of motion for $\theta$ :

$\frac{d}{dt} \frac{\partial{L}}{\partial{\dot{\theta}}} = \frac{\partial{L}}{\partial{\theta}}$

Evaluating gives:

$\ddot{\theta} ( \frac{m L^2}{3} + ML^2) = - \frac{1}{2} \, mg \, L - Mg \, L - k L^2 \theta$

Homogeneous equation:

$\ddot{\theta} \big( \frac{m L^2}{3} + ML^2\big) = -k L^2 \theta \\ \ddot{\theta} \big( m + 3M \big) = -3k \theta$

This corresponds to harmonic motion with the following angular frequency:

$\omega = \dot{\theta} = \sqrt{\frac{3k}{m + 3M}}$

Relevant wiki: Simple Harmonic Motion - Problem Solving

Consider small angles $\sin{\theta} \simeq \theta$ , $\cos{\theta} \simeq 1$ .

Force of the spring is $k \theta L$ . Force times distance $L$ from the pivot gives torque equal to $\theta k L^2$ .

Gravitational forces are $gm$ and $gM$ which gives torques equal to $gm\frac{L}{2}$ and $gML$ .

From Newton's 2nd law:

$\ddot{\theta} I = - \theta k L^2 - gL(\frac{m}{2}+M).$

One can divide by planks moment of inertia $I = ML^2 + \frac{1}{3}mL^2$ which gives:

$\ddot{\theta} = - \theta \frac{k L^2}{I} - \frac{gL}{I}(\frac{m}{2}+M).$

From harmonic oscillator equation:

$\omega^2 = \frac{k L^2}{I} = \frac{k L^2}{ML^2 + \frac{1}{3}mL^2} = \frac{3k}{3M + m}.$

It is worth noting that we cannot simultaneously have the equilibrium position of the plank being horizontal and the natural length of the spring being the height of the pivoted plank. The spring must be under compression at the horizontal for the plank to be in equilibrium there - the thrust in the spring has to oppose the weights of the plank and the mass.