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| Solutions of Savchenko Problems in Physics <br> | | Solutions of Savchenko Problems in Physics <br> |
| <i><b>knowledge must be free</b></i> | | <i><b>knowledge must be free</b></i> |
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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> |
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| $3.1.5.$ The length of the string of a mathematical pendulum is $l$, the mass of the ball is $m$. Determine the force acting on the ball as it deviates from the equilibrium position by $x$ in the case where $x \ll l$. How does the potential energy of the ball depend on $x$? | | $3.1.5.$ The length of the string of a mathematical pendulum is $l$, the mass of the ball is $m$. Determine the force acting on the ball as it deviates from the equilibrium position by $x$ in the case where $x \ll l$. How does the potential energy of the ball depend on $x$? |
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| For problem $3.1.5$ | | For problem $3.1.5$ |
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| <h3>Solution</h3> | | <h3>Solution</h3> |
| <p> | | <p> |
| When the pendulum swings away, it is acted upon by a restoring force $F$ acting in the direction opposite to the direction of the velocity, which is restoring and equal to | | When the pendulum swings away, it is acted upon by a restoring force $F$ acting in the direction opposite to the direction of the velocity, which is restoring and equal to |
| $$ OX:F=-mg\sin\varphi=\fbox{$-mg\frac{x}{l}$} $$ | | $$ OX:F=-mg\sin\varphi=\fbox{$-mg\frac{x}{l}$} $$ |
| <p style="text-align: right; font-style: italic; font-size: 14;"> | | <p style="text-align: right; font-style: italic; font-size: 14;"> |
| Aliaksandr Melnichenka | | Aliaksandr Melnichenka |
| </p> | | </p> |
| <h3>Alternative solution</h3> | | <h3>Alternative solution</h3> |
| <p> | | <p> |
| Alternatively, one could write the change in energy as an integral of the work done on the interval $dx$: | | Alternatively, one could write the change in energy as an integral of the work done on the interval $dx$: |
| $$ U=\int_{0}^{U}dU=\int_{0}^{x}mg\frac{x}{l}dx=\boxed{\frac{mgx^2}{2l}} $$ | | $$ U=\int_{0}^{U}dU=\int_{0}^{x}mg\frac{x}{l}dx=\boxed{\frac{mgx^2}{2l}} $$ |
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| <p style="text-align: right; font-style: italic; font-size: 14;"> | | <p style="text-align: right; font-style: italic; font-size: 14;"> |
| Dzikan Mikita<br> | | Dzikan Mikita<br> |
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| <h4>Answer</h4> | | <h4>Answer</h4> |
| <p> | | <p> |
| $$F=-mg\frac{x}{l};\quad U=\frac{mgx^2}{2l}$$ | | $$F=-mg\frac{x}{l};\quad U=\frac{mgx^2}{2l}$$ |
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