Added 2.4.20 3.3.8

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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="../#3.3">$\leftarrow$Back</a></h3>
+
+ <h3> Statement </h3>
+ <p>
+ $3.3.8^*.$ The balloon deforms when it weakly hits the wall, as shown in the figure. The maximum deformation of the balloon $x$ is much smaller than its radius $R$. Ignoring the change in excess pressure $\Delta p$ of air in the balloon and the elasticity of the shell, estimate the time of impact with the wall. The mass of the ball $m$.
+
+</p>
+<center>
+ <figure>
+ <img src="statement.png"
+ loading="lazy" width="150" />
+ <figcaption>
+ For problem $3.3.8^*$
+ </figcaption>
+ </figure>
+</center>
+<p>
+ </p>
+
+ <h3>Solution</h3>
+ <p>
+
+</p>
+<center>
+ <figure>
+ <img src="draw.png"
+ loading="lazy" width="180" />
+ <figcaption>
+ Geometry of the problem
+ </figcaption>
+ </figure>
+</center>
+<p>
+
+$$y=R-x$$
+$$r^2=R^2-(R-x)^2$$
+$$S=\pi r^2=\pi (R^2-(R-x)^2)$$
+Considering $x \ll R$
+$$S=\pi r^2=\pi (R^2-(R-x)^2)=2\pi Rx$$
+Newton's second law in differential form:
+$$-F=m\ddot{x}$$
+$$m\ddot{x}+F=0 \Rightarrow m\ddot{x}+2\pi Rx \cdot \Delta p=0$$
+$$\ddot{x}+\frac{2\pi R\Delta p}{m}x=0$$
+Recalling the differential equation of harmonic motion
+$$\ddot{x}+\omega ^2x=0$$
+Find the cyclic frequency $\omega$ of oscillation
+$$\omega=\sqrt{\frac{2\pi R\Delta p}{m}}$$
+Since we're interested in half of the full period, estimated time of impact with the wall could be found as
+$$t=\frac{1}{2}\frac{2\pi}{\omega}=\frac{\pi}{\omega}$$
+$$\boxed{t=\pi\sqrt{\frac{m}{2\pi R\Delta p}}}$$
+ </p>
+
+ <h4>Answer</h4>
+ <p>
+ $$t=\pi\sqrt{m/(2\pi R\Delta p)}$$
+ </p>
+ <p style="text-align: right; font-style: italic; font-size: 14;">
+ Yuldashev Ulugbek<br>
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