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Jorge Llorente 08 правка от
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+### Statement
+
+$3.1.13.$ [Insert the problem statement]
+
+### Solution
+
+![For problem $3.1.13$ |1054x697, 31%](../../img/3.1.13/Imagen1.png)
+
+According to Problems~3.1.12 and~3.1.13, we consider the following configuration.
+
+A small charged body of mass $m$ can slide freely on a vertical spoke. At the lower end of the spoke there is a fixed charge of the same sign as that of the body. The equilibrium position of the body is located at a distance $R$ from the fixed charge. A small additional mass $\Delta m$ is then rigidly attached to the charged body.
+
+This modification alters the equilibrium distance. Initially, the balance of forces gives
+\begin{equation}
+\frac{k Q q}{r^2} = mg,
+\end{equation}
+from which
+\begin{equation}
+r = \sqrt{\frac{k Q q}{m g}}.
+\end{equation}
+
+After adding the mass $\Delta m$, the equilibrium condition becomes
+\begin{equation}
+\frac{k Q q}{r'^2} = (m + \Delta m) g,
+\end{equation}
+leading to
+\begin{equation}
+r' = \sqrt{\frac{k Q q}{(m + \Delta m) g}}.
+\end{equation}
+
+The simplest way to solve the problem is by using the principle of conservation of mechanical energy, since the system is conservative:
+\begin{equation}
+\Delta \Sigma E = 0.
+\end{equation}
+
+Thus,
+\begin{equation}
+E_{\text{el}}(r) + E_{\text{grav}}(r)
+=
+E_{\text{el}}(r') + E_{\text{grav}}(r') + E_{\text{kin}}.
+\end{equation}
+
+Expanding each term, we obtain
+\begin{equation}
+(m+\Delta m) g r + \frac{k Q q}{r}
+=
+(m+\Delta m) g r' + \frac{k Q q}{r'} + \frac{1}{2}(m+\Delta m) v^2.
+\end{equation}
+
+Substituting the expression
+\[
+r' = \sqrt{\frac{k Q q}{(m + \Delta m) g}},
+\]
+the equation becomes
+\begin{equation}
+(m+\Delta m) g r + \frac{k Q q}{r}
+=
+(m+\Delta m) g \sqrt{\frac{k Q q}{(m + \Delta m) g}}
++
+\frac{k Q q}{\sqrt{\frac{k Q q}{(m + \Delta m) g}}}
++
+\frac{1}{2}(m+\Delta m) v^2.
+\end{equation}
+
+Solving for the velocity, the maximum speed is found to be
+\begin{equation}
+v_{\max}
+=
+\sqrt{
+2 g R
+\left(
+1 + \frac{m}{m+\Delta m}
+- 2 \sqrt{\frac{m}{m+\Delta m}}
+\right)
+}.
+\end{equation}
+
+This result can also be justified by analyzing the motion itself. Since no dissipative forces (such as air resistance) are present, the system is conservative, and the motion is therefore simple harmonic motion (SHM).
+
+For SHM, the velocity is given by
+\begin{equation}
+v = \omega \sqrt{A^2 - x^2},
+\end{equation}
+where $A$ is the amplitude and $\omega$ the angular frequency. The velocity reaches its maximum value at the equilibrium position, where $x = 0$, yielding
+\begin{equation}
+v_{\max} = A \omega.
+\end{equation}
+
+The amplitude corresponds to the difference between the original equilibrium position and the new one:
+\begin{equation}
+A = r - r'
+= \sqrt{\frac{k Q q}{m g}}
+-
+\sqrt{\frac{k Q q}{(m + \Delta m) g}}.
+\end{equation}
+
+Substituting this expression into the SHM formulas yields an equivalent result to that obtained via energy conservation. Although this approach is equally valid, it requires a Taylor expansion, which unnecessarily complicates the analysis.
+
+
+
+#### Answer
+
+[Insert a concise answer or boxed result]