Translated 3.1.1-3.1.17

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+ <span><img src = "../../img/book.png"><span><span>Savchenko Solutions</span>
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+ Solutions&nbsp;of&nbsp;Savchenko Problems&nbsp;in&nbsp;Physics <br>
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+ <h3 id="back-link"><a href="../../#3.1">$\leftarrow$Back</a></h3>
+
+ <h3> Statement </h3>
+ <p>
+ $3.1.7.$ A horizontal trough to the left of the bottom line is bent on a cylindrical surface of radius $r$, and to the right, on a surface of radius $R$. Determine the ratio of the largest deflections to the left and to the right during small oscillations of the body in this trough.
+</p>
+<center>
+ <figure>
+ <img src="https://savchenkosolutions.com/3/3.1.7/statement.png"
+ loading="lazy" width="150" />
+ <figcaption>
+ For problem $3.1.7$
+ </figcaption>
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+<p>
+ </p>
+
+ <h3>Solution</h3>
+ <p>
+ According to the law of conservation of body energy, the maximum potential energy will be conserved in both cases
+$$ E_{p1}=E_{p2} $$
+And therefore the height $h$ will rise relative to the lower point of the trajectory to which the body will rise
+will be equal
+
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+<img src="https://savchenkosolutions.com/3/3.1.7/3.1.7_1.png"
+loading="lazy" width="230" />
+<figcaption>
+The position of the ball at the extreme points of the trajectory
+</figcaption>
+</figure>
+</center>
+<p>
+We will use the expression $\Delta U$ for small oscillations obtained in <a href="../3.1.5">3.1.5</a>
+$$ \Delta U = \frac{mgx_1^2}{2r}=\frac{mgx_2^2}{2R} $$
+Whence it follows that
+$$ \boxed{\frac{x_2}{x_1}=\sqrt{\frac{R}{r}}} $$
+
+ </p>
+
+ <h4>Answer</h4>
+ <p>
+ $$\frac{x_2}{x_1}=\sqrt{\frac{R}{r}}$$
+ </p>
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