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			<h3 id="back-link"><a href="../../#3.3">$\leftarrow$Back</a></h3>

			<h3> Statement </h3>
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			$3.3.8^*.$ 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 $m$.

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			<img src="statement.png"
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			For problem $3.3.8^*$
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			<h3>Solution</h3>
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			loading="lazy" width="180" />
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			Geometry of the problem
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			<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:
			$$-F=m\ddot{x}$$
		@@ -90,9 +90,9 @@
			$$m\ddot{x}+F=0 \Rightarrow m\ddot{x}+2\pi Rx \cdot \Delta p=0$$
			$$\ddot{x}+\frac{2\pi R\Delta p}{m}x=0$$
			Recalling the differential equation of harmonic motion
		−	$$\ddot{x}+\omega ^2x=0$$
		+	$$\ddot{x}+\omega^2x=0$$
			Find the cyclic frequency $\omega$ of oscillation
		−	$$\omega=\sqrt{\frac{2\pi R\Delta p}{m}}$$
		+	$$\omega =\sqrt{\frac{2\pi R\Delta p}{m}}$$
			Since we're interested in half of the full period, estimated time of impact with the wall could be found as
			$$t=\frac{1}{2}\frac{2\pi}{\omega}=\frac{\pi}{\omega}$$
			$$\boxed{t=\pi\sqrt{\frac{m}{2\pi R\Delta p}}}$$
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			<h4>Answer</h4>
			<p>
			$$t=\pi\sqrt{m/(2\pi R\Delta p)}$$
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	<h3> Statement </h3>		<h3> Statement </h3>
	<p>		<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 $m$.		$3.3.8^*.$ 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>		</p>
	<center>		<center>
	<figure>		<figure>
	<img src="statement.png"		<img src="statement.png"
	loading="lazy" width="150" />		loading="lazy" width="150" />
	<figcaption>		<figcaption>
	For problem $3.3.8^*$		For problem $3.3.8^*$
	</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="draw.png"		<img src="draw.png"
	loading="lazy" width="180" />		loading="lazy" width="180" />
	<figcaption>		<figcaption>
	Geometry of the problem		Geometry of the problem
	</figcaption>		</figcaption>
	</figure>		</figure>
	</center>		</center>
	<p>		<p>

	$$y=R-x$$		$$y=R-x$$
	$$r^2=R^2-(R-x)^2$$		$$r^2=R^2-(R-x)^2$$
	$$S=\pi r^2=\pi (R^2-(R-x)^2)$$		$$S=\pi r^2=\pi (R^2-(R-x)^2)$$
	Considering $x \ll R$		Considering $x \ll R$
	$$S=\pi r^2=\pi (R^2-(R-x)^2)=2\pi Rx$$		$$S=\pi r^2=\pi (R^2-(R-x)^2)=2\pi Rx$$
	Newton's second law in differential form:		Newton's second law in differential form:
	$$-F=m\ddot{x}$$		$$-F=m\ddot{x}$$
	@@ -90,9 +90,9 @@
	$$m\ddot{x}+F=0 \Rightarrow m\ddot{x}+2\pi Rx \cdot \Delta p=0$$		$$m\ddot{x}+F=0 \Rightarrow m\ddot{x}+2\pi Rx \cdot \Delta p=0$$
	$$\ddot{x}+\frac{2\pi R\Delta p}{m}x=0$$		$$\ddot{x}+\frac{2\pi R\Delta p}{m}x=0$$
	Recalling the differential equation of harmonic motion		Recalling the differential equation of harmonic motion
	$$\ddot{x}+\omega ^2x=0$$		$$\ddot{x}+\omega^2x=0$$
	Find the cyclic frequency $\omega$ of oscillation		Find the cyclic frequency $\omega$ of oscillation
	$$\omega=\sqrt{\frac{2\pi R\Delta p}{m}}$$		$$\omega =\sqrt{\frac{2\pi R\Delta p}{m}}$$
	Since we're interested in half of the full period, estimated time of impact with the wall could be found as		Since we're interested in half of the full period, estimated time of impact with the wall could be found as
	$$t=\frac{1}{2}\frac{2\pi}{\omega}=\frac{\pi}{\omega}$$		$$t=\frac{1}{2}\frac{2\pi}{\omega}=\frac{\pi}{\omega}$$
	$$\boxed{t=\pi\sqrt{\frac{m}{2\pi R\Delta p}}}$$		$$\boxed{t=\pi\sqrt{\frac{m}{2\pi R\Delta p}}}$$
	</p>		</p>

	<h4>Answer</h4>		<h4>Answer</h4>
	<p>		<p>
	$$t=\pi\sqrt{m/(2\pi R\Delta p)}$$		$$t=\pi\sqrt{m/(2\pi R\Delta p)}$$
	</p>		</p>


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