Edit to “Solution”

Valter edited Aug 11, 2026, 6:18 PM

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			### Statement

			$3.1.1.$ A ball is fixed in the middle of a string of length $2$l. What is the total force acting on the ball from the side of the string if its transverse displacement from the equilibrium position is $x \ll $, and the string tension force $F$ does not depend on the displacement? Why, for small displacements, can we consider the dependence of the force acting on the ball on $x$ to be linear? What is the direction of this force with respect to the displacement? Find how the potential energy of the ball depends on the small displacement $x$. What is the velocity of the ball when it passes its equilibrium position if its maximum displacement is $x_0$? The mass of the ball is $m$.

			![ For problem $3.1.1$ \|581x164, 51%](../../img/3.1.1/3.1.1.png)

			### Solution

			In addition to the force of gravity $mg$, the ball is subject to the force of tension of the string $F$, which has a vertical component $F'$:

			$$
			F'=-2F\cos\alpha =-2F\frac{x}{\sqrt{l^2+x^2}}\tag{1}
			$$

			Since the oscillations are small, we take into account that $x\ll l$

			$$
			\frac{x}{\sqrt{l^2+x^2}} = \frac{1}{\sqrt{1+\frac{l^2}{x^2}}}
			$$

			As $x\ll l$, то $\frac{l^2}{x^2} \gg 1$:

			$$
			\frac{1}{\sqrt{1+\frac{l^2}{x^2}}}=\frac{1}{\sqrt{\frac{l^2}{x^2}}}=\frac{x}{l}
			$$

			Substituting into $(1)$

			$$
			\boxed{F'=-2F\frac{x}{l}}
			$$

		@@ -33,7 +33,7 @@Solution
			We find the small change in energy as

			$$
		−	dU=2F\cos\alpha dx=2F~~\sqrt{~~\frac{x}{l^2+x^2}}=2F\frac{x}{l}dx
		+	dU=2F\\cos\\alpha dx=2F\\frac{x}{\\sqrt{l^2+x^2}}dx=2F\\frac{x}{l}dx
			$$

			Next, we integrate $dU$ from $0$ to $U$

			$$
			U=\int_{0}^{U}dU=\int_{0}^{x}2F\frac{x}{l}dx=\boxed{\frac{Fx^2}{l}}
			$$

			From the law of conservation of energy, we obtain the desired speed at maximum displacement

			$$
			\frac{Fx_0^2}{l}=\frac{m\upsilon^2}{2}\Rightarrow \boxed{\upsilon =x_0\sqrt{\frac{2F}{ml}}}
			$$

			#### Answer

			$$
			\upsilon =x_0\sqrt{\frac{2F}{ml}}
			$$






		unchanged lines 22

	### Statement		### Statement

	$3.1.1.$ A ball is fixed in the middle of a string of length $2$l. What is the total force acting on the ball from the side of the string if its transverse displacement from the equilibrium position is $x \ll $, and the string tension force $F$ does not depend on the displacement? Why, for small displacements, can we consider the dependence of the force acting on the ball on $x$ to be linear? What is the direction of this force with respect to the displacement? Find how the potential energy of the ball depends on the small displacement $x$. What is the velocity of the ball when it passes its equilibrium position if its maximum displacement is $x_0$? The mass of the ball is $m$.		$3.1.1.$ A ball is fixed in the middle of a string of length $2$l. What is the total force acting on the ball from the side of the string if its transverse displacement from the equilibrium position is $x \ll $, and the string tension force $F$ does not depend on the displacement? Why, for small displacements, can we consider the dependence of the force acting on the ball on $x$ to be linear? What is the direction of this force with respect to the displacement? Find how the potential energy of the ball depends on the small displacement $x$. What is the velocity of the ball when it passes its equilibrium position if its maximum displacement is $x_0$? The mass of the ball is $m$.

	![ For problem $3.1.1$ \|581x164, 51%](../../img/3.1.1/3.1.1.png)		![ For problem $3.1.1$ \|581x164, 51%](../../img/3.1.1/3.1.1.png)

	### Solution		### Solution

	In addition to the force of gravity $mg$, the ball is subject to the force of tension of the string $F$, which has a vertical component $F'$:		In addition to the force of gravity $mg$, the ball is subject to the force of tension of the string $F$, which has a vertical component $F'$:

	$$		$$
	F'=-2F\cos\alpha =-2F\frac{x}{\sqrt{l^2+x^2}}\tag{1}		F'=-2F\cos\alpha =-2F\frac{x}{\sqrt{l^2+x^2}}\tag{1}
	$$		$$

	Since the oscillations are small, we take into account that $x\ll l$		Since the oscillations are small, we take into account that $x\ll l$

	$$		$$
	\frac{x}{\sqrt{l^2+x^2}} = \frac{1}{\sqrt{1+\frac{l^2}{x^2}}}		\frac{x}{\sqrt{l^2+x^2}} = \frac{1}{\sqrt{1+\frac{l^2}{x^2}}}
	$$		$$

	As $x\ll l$, то $\frac{l^2}{x^2} \gg 1$:		As $x\ll l$, то $\frac{l^2}{x^2} \gg 1$:

	$$		$$
	\frac{1}{\sqrt{1+\frac{l^2}{x^2}}}=\frac{1}{\sqrt{\frac{l^2}{x^2}}}=\frac{x}{l}		\frac{1}{\sqrt{1+\frac{l^2}{x^2}}}=\frac{1}{\sqrt{\frac{l^2}{x^2}}}=\frac{x}{l}
	$$		$$

	Substituting into $(1)$		Substituting into $(1)$

	$$		$$
	\boxed{F'=-2F\frac{x}{l}}		\boxed{F'=-2F\frac{x}{l}}
	$$		$$

	@@ -33,7 +33,7 @@Solution
	We find the small change in energy as		We find the small change in energy as

	$$		$$
	dU=2F\cos\alpha dx=2F~~\sqrt{~~\frac{x}{l^2+x^2}}=2F\frac{x}{l}dx		dU=2F\\cos\\alpha dx=2F\\frac{x}{\\sqrt{l^2+x^2}}dx=2F\\frac{x}{l}dx
	$$		$$

	Next, we integrate $dU$ from $0$ to $U$		Next, we integrate $dU$ from $0$ to $U$

	$$		$$
	U=\int_{0}^{U}dU=\int_{0}^{x}2F\frac{x}{l}dx=\boxed{\frac{Fx^2}{l}}		U=\int_{0}^{U}dU=\int_{0}^{x}2F\frac{x}{l}dx=\boxed{\frac{Fx^2}{l}}
	$$		$$

	From the law of conservation of energy, we obtain the desired speed at maximum displacement		From the law of conservation of energy, we obtain the desired speed at maximum displacement

	$$		$$
	\frac{Fx_0^2}{l}=\frac{m\upsilon^2}{2}\Rightarrow \boxed{\upsilon =x_0\sqrt{\frac{2F}{ml}}}		\frac{Fx_0^2}{l}=\frac{m\upsilon^2}{2}\Rightarrow \boxed{\upsilon =x_0\sqrt{\frac{2F}{ml}}}
	$$		$$

	#### Answer		#### Answer

	$$		$$
	\upsilon =x_0\sqrt{\frac{2F}{ml}}		\upsilon =x_0\sqrt{\frac{2F}{ml}}
	$$		$$






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