Two beads of mass m each, connected to each other by a spring of stiffness k, are held on smooth rods rigidly fixed in the wall. The spring is stretched and its length is l. The distance between the free ends of the rods is equal to the length of the undeformed spring. The beads are released. How fast will the spring move in the x-direction after the beads jump off the rod? What will be the greatest compression strain of the spring?

Solutions of Savchenko Problems in Physics

Aliaksandr Melnichenka
October 2023

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    <h3> Statement </h3>
    <p>
        $2.4.16.$ Two beads of mass $m$ each, connected to each other by a spring of stiffness $k$, are held on smooth rods rigidly fixed in the wall. The spring is stretched and its length is $l$. The distance between the free ends of the rods is equal to the length of the undeformed spring. The beads are released. How fast will the spring move in the $x$-direction after the beads jump off the rod? What will be the greatest compression strain of the spring?

    <h3>Solution</h3>
    <p>
        Preservation of energy at the initial stage

$$\frac{k\left(l-l_0\right)^2}{2} = 2\frac{mv^2}{2}$$
From where
$$v = \sqrt{\frac{k}{2m}}\left(l-l_0\right)$$
The greatest compression will be when it has nowhere to compress $(v_y=0)$
Alternative for horizontal projection
$$v_x=v\cos\alpha;v_x=\text{const}$$
After substituting
$$\boxed{v_x = \sqrt{\frac{k}{2m}}\left(l-l_0\right)\cos\alpha}$$
Conservation of energy with $x_\text{max}$ deformation
$$\frac{k\left(l-l_0\right)^2}{2} = 2\frac{mv_x^2}{2} + \frac{kx_\text{max}^2}{2}$$
From where
$$x_\text{max}^2=\left(l-l_0\right)^2\left(1-\cos^2\alpha\right)$$
After using pythagorean trigonometric identity:
$$\boxed{x_\text{max} = \left(l-l_0\right) \sin \alpha}$$

    <h4>Answer</h4>
    <p>
        $$\begin{aligned}v_x=(l-l_0)\sqrt{k/(2m)} \cos\alpha; x=(l-l_0)\sin\alpha.\end{aligned}$$
    </p>
    <p style="text-align: right; font-style: italic; font-size: 14;">
      Almaskhan Arsen<br>
    </p>

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