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| <title>A ball is fixed in the middle of a string of length 2l. 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.</title> | | <title>A ball is fixed in the middle of a string of length 2l. 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.</title> |
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| <h3 id="back-link"><a href="../../#3.1">$\leftarrow$Back</a></h3> | | <h3 id="back-link"><a href="../../#3.1">$\leftarrow$Back</a></h3> |
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| <h3> Statement </h3> | | <h3> Statement </h3> |
| <p> | | <p> |
| $3.1.1.$ 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 $m$. |
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| loading="lazy" width="300" /> | | loading="lazy" width="300" /> |
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| For problem $3.1.1$ | | For problem $3.1.1$ |
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| <h3>Solution</h3> | | <h3>Solution</h3> |
| <p> | | <p> |
| 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 $F'$: |
| $$ F'=-2Fcos\alpha=-2F\frac{x}{\sqrt{l^2+x^2}}\quad(1) $$ | | $$ F'=-2Fcos\alpha=-2F\frac{x}{\sqrt{l^2+x^2}}\quad(1) $$ |
| Since the oscillations are small, we take into account that $x\ll l$ | | Since the oscillations are small, we take into account that $x\ll l$ |
| $$ \frac{x}{\sqrt{l^2+x^2}} = \frac{1}{\sqrt{1+\frac{l^2}{x^2}}} $$ | | $$ \frac{x}{\sqrt{l^2+x^2}} = \frac{1}{\sqrt{1+\frac{l^2}{x^2}}} $$ |
| Т.к. $x\ll l$, то $\frac{l^2}{x^2} \gg 1$: | | Т.к. $x\ll l$, 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 $(1)$ |
| $$ \boxed{F'=-2F\frac{x}{l}} $$ | | $$ \boxed{F'=-2F\frac{x}{l}} $$ |
| We find the small change in energy as | | We find the small change in energy as |
| $$ dU=2Fcos\alpha dx=2F\sqrt{\frac{x}{l^2+x^2}}=2F\frac{x}{l}dx $$ | | $$ dU=2Fcos\alpha dx=2F\sqrt{\frac{x}{l^2+x^2}}=2F\frac{x}{l}dx $$ |
| 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 |
| $$ \frac{Fx_0^2}{l}=\frac{m\upsilon^2}{2}\Rightarrow \boxed{\upsilon=x_0\sqrt{\frac{2F}{ml}}} $$ | | $$ \frac{Fx_0^2}{l}=\frac{m\upsilon^2}{2}\Rightarrow \boxed{\upsilon=x_0\sqrt{\frac{2F}{ml}}} $$ |
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| <h4>Answer</h4> | | <h4>Answer</h4> |
| <p> | | <p> |
| $$\upsilon=x_0\sqrt{\frac{2F}{ml}}$$ | | $$\upsilon=x_0\sqrt{\frac{2F}{ml}}$$ |
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