Answer:
[tex]T = 2\pi\cdot \sqrt{\frac{l}{g + a} }[/tex]
Explanation:
It is known that stick is experimenting a Simple Harmonic Movement and, to be exactly, can be modelled as a simple pendulum. The period of oscilation of the stick is:
[tex]T = \frac{2\pi}{\omega}[/tex]
The pendulum is modelled by the Newton's Laws. The Free Body Diagram is presented below:
[tex]\Sigma F_{r} = T - m\cdot g \cdot \cos \theta = m\cdot (\omega^{2}\cdot l + a\cdot \cos \theta)[/tex]
[tex]\Sigma F_{t} = m\cdot g \cdot \sin \theta = m\cdot (l\cdot \alpha - a \cdot \sin \theta)[/tex]
Let assume that pendulum is just experimenting small oscillations, so that:
[tex]\theta \approx \sin \theta[/tex]
Then:
[tex]m\cdot g \cdot \theta = m\cdot (l\cdot \alpha - a\cdot \theta)[/tex]
[tex]g\cdot \theta = l\cdot \alpha - a\cdot \theta[/tex]
[tex](g + a)\cdot \theta = l\cdot \alpha[/tex]
[tex]\alpha = \frac{g+a}{l}\cdot \theta[/tex]
Where [tex]\omega =\sqrt{\frac{g + a}{l} }[/tex].
Finally, the period is:
[tex]T = 2\pi\cdot \sqrt{\frac{l}{g + a} }[/tex]
