Savchenko Solutions

Solutions of Savchenko Physics Textbook

Aliaksandr Melnichenka
October 2023

    <h3 id="back-link"><a href="/#2.3">$\leftarrow$Back</a></h3>

    <h3> Statement </h3>
    <p>
        $2.3.44.$ An unstretched rubber cord of length $2l$ is attached to the walls by its ends. A weight of mass $m$ was attached to the middle of the cord, which was then released without a push. When vibrations occur, the greatest distance to which the load is lowered is $x_0$. What is the stiffness of this cord?

    <h3>Solution</h3>
    <p>
        

Potential energy of the weight: $$E_p = -mgx_0$$ Potential energy of rubber cord $$E_p' = k \left( \frac{\sqrt{l^2 + x_0^2} - l}{2} \right)^2$$ From the law of conservation of energy, the total energy of the system must equal zero $$E = 0 = E_p + E_p' + E_p'$$ It follows from the figure that the stretching of the cord is $$\Delta L = \sqrt{l^2 + x_0^2} - l$$

Now, considering the system as a single spring, we use superposition:
$$\frac{1}{k} = \frac{1}{k'} + \frac{1}{k'} \implies k' = 2k$$
From where stiffness of the cord could be found as
$$mgx_0 = 8 \cdot 2k \cdot \left( \frac{\sqrt{l^2 + x_0^2} - l}{2} \right)^2$$

$$\boxed{k = \frac{mgx_0}{2 \cdot \left( \sqrt{l^2 + x_0^2} - l \right)^2}}$$

    <h4>Answer</h4>
    <p>
        $$k = \frac{mgx_0}{2 \cdot \left( \sqrt{l^2 + x_0^2} - l \right)^2}$$
    </p>
    <p style="text-align: right; font-style: italic; font-size: 14;">   
      Almaskhan Arsen<br>
    </p>


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