$2.3.44.$ 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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<img src="statement.png"
loading="lazy" width="300" />
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For problem $2.3.44$
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<h3>Solution</h3>
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<img src="2.3.44.jpg"
loading="lazy" width="300" />
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Geometry of the problem
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<p>
@@ -76,16 +76,16 @@
Potential energy of the weight:
$$E_p = -mgx_0 $$
Potential energy of rubber cord
−
$$E_p' = k \left( \frac{\sqrt{l^2 + x_0^2} - l}{2}\right)^2$$
+
$$E_p' = k'\frac{(\sqrt{l^2 + x_0^2} - l)^2}{2}$$
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:
+
Since we now consider them as separate springs, cords, 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
<small>All rights belong to the authors. <br> Commercial use of materials - with the written permission of the authors. <br> astrosander01@gmail.com <br></small>
$2.3.44.$ 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?
$2.3.44.$ 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?
</p>
</p>
<center>
<center>
<figure>
<figure>
<img src="statement.png"
<img src="statement.png"
loading="lazy" width="300" />
loading="lazy" width="300" />
<figcaption>
<figcaption>
For problem $2.3.44$
For problem $2.3.44$
</figcaption>
</figcaption>
</figure>
</figure>
</center>
</center>
<p>
<p>
</p>
</p>
<h3>Solution</h3>
<h3>Solution</h3>
<p>
<p>
</p>
</p>
<center>
<center>
<figure>
<figure>
<img src="2.3.44.jpg"
<img src="2.3.44.jpg"
loading="lazy" width="300" />
loading="lazy" width="300" />
<figcaption>
<figcaption>
Geometry of the problem
Geometry of the problem
</figcaption>
</figcaption>
</figure>
</figure>
</center>
</center>
<p>
<p>
@@ -76,16 +76,16 @@
Potential energy of the weight:
Potential energy of the weight:
$$E_p = -mgx_0 $$
$$E_p = -mgx_0 $$
Potential energy of rubber cord
Potential energy of rubber cord
$$E_p' = k \left( \frac{\sqrt{l^2 + x_0^2} - l}{2}\right)^2$$
$$E_p' = k'\frac{(\sqrt{l^2 + x_0^2} - l)^2}{2}$$
From the law of conservation of energy, the total energy of the system must equal zero
From the law of conservation of energy, the total energy of the system must equal zero
$$E = 0 = E_p + E_p' + E_p'$$
$$E = 0 = E_p + E_p' + E_p'$$
It follows from the figure that the stretching of the cord is
It follows from the figure that the stretching of the cord is
$$\Delta L = \sqrt{l^2 + x_0^2} - l$$
$$\Delta L = \sqrt{l^2 + x_0^2} - l$$
Now, considering the system as a single spring, we use superposition:
Since we now consider them as separate springs, cords, we use superposition:
$$\frac{1}{k} = \frac{1}{k'} + \frac{1}{k'}\implies k' = 2k $$
$$\frac{1}{k} = \frac{1}{k'} + \frac{1}{k'}\implies k' = 2k $$
From where stiffness of the cord could be found as
From where stiffness of the cord could be found as
<small>All rights belong to the authors. <br> Commercial use of materials - with the written permission of the authors. <br> astrosander01@gmail.com <br></small>
<small>All rights belong to the authors. <br> Commercial use of materials - with the written permission of the authors. <br> astrosander01@gmail.com <br></small>
