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+ <h2>Solutions of Savchenko Physics Textbook</h2>
+ <p class="author">
+ Aliaksandr Melnichenka <br/>
+ October 2023
+ </p>
+ </header>
+
+ <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?
+
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+<center>
+ <figure>
+ <img src="statement.png"
+ loading="lazy" width="300" />
+ <figcaption>
+ For problem $2.3.44$
+ </figcaption>
+ </figure>
+</center>
+<p>
+ </p>
+
+ <h3>Solution</h3>
+ <p>
+
+</p>
+<center>
+ <figure>
+ <img src="2.3.44.jpg"
+ loading="lazy" width="300" />
+ <figcaption>
+ Geometry of the problem
+ </figcaption>
+ </figure>
+</center>
+<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
+$$mgx_0 = 8 \cdot 2k \cdot \left( \frac{\sqrt{l^2 + x_0^2} - l}{2} \right)^2$$
+
+$$\boxed{k = \frac{mgx_0}{2 \cdot \left( \sqrt{l^2 + x_0^2} - l \right)^2}}$$
+ </p>
+
+ <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>
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