8. (15 points) A structure consisting of a rigidly connected light rod and a small load with mass \( m = 1.6 \) kg can oscillate under the action of two springs with stiffnesses \( k_{1} = 10 \frac{\mathrm{H}}{\mathrm{M}} \) and \( k_{2} = 7.5 \frac{\mathrm{H}}{\mathrm{M}} \), moving during rotation without friction around a vertical axis \( O \) on a smooth horizontal table surface. The springs are light, their axes are horizontal, and the points of attachment to the rod divide it into three equal parts. In the equilibrium position, the springs are undeformed. Find the period of small oscillations of the structure.

![](https://cdn.mathpix.com/cropped/2024_05_06_be79a21ed2ca3a6294abg-12.jpg?height=437&width=965&top_left_y=712&top_left_x=617)
Answer: $\approx 3.8 s$

Solution. Let's deflect the rod from its equilibrium position by a small angle $\alpha$. In this position, the total mechanical energy of the system is:

$\frac{k_{1} x_{1}^{2}}{2}+\frac{k_{2} x_{2}^{2}}{2}+\frac{m v^{2}}{2}=$ const

where $x_{1}=\alpha \frac{l}{3}, x_{2}=\alpha \frac{2 l}{3}, v=\omega l$.

Differentiate the law of conservation of total mechanical energy with respect to time: $\frac{k_{1}}{2} \cdot \frac{l^{2}}{9} \cdot 2 \alpha \cdot \omega+\frac{k_{2}}{2} \cdot \frac{4 l^{2}}{9} \cdot 2 \alpha \cdot \omega+\frac{m}{2} \cdot l^{2} \cdot 2 \omega \cdot \varepsilon=0$,

$\frac{k_{1}+4 k_{2}}{9 m} \cdot \alpha+\varepsilon=0$.

As a result, we obtain that the cyclic frequency of oscillations is:

$\omega_{0}=\sqrt{\frac{k_{1}+4 k_{2}}{9 m}}$.

The period of oscillations of the structure: $T=\frac{2 \pi}{\omega_{0}}=6 \pi \sqrt{\frac{m}{k_{1}+4 k_{2}}}=1.2 \pi \approx 3.8 s$.(2 points)