Translated 3.1.1-3.1.17
en/3.1.5.md
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| + | <title>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?</title> | ||
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| + | <span><img src = "../../img/book.png"><span><span>Savchenko Solutions</span> | ||
| + | </div> | ||
| + | <p class="author"> | ||
| + | Solutions of Savchenko Problems in Physics <br> | ||
| + | <i><b>knowledge must be free</b></i> | ||
| + | </p> | ||
| + | </header> | ||
| + | |||
| + | <h3 id="back-link"><a href="../../#3.1">$\leftarrow$Back</a></h3> | ||
| + | |||
| + | <h3> Statement </h3> | ||
| + | <p> | ||
| + | $3.1.5.$ 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 $x$? | ||
| + | </p> | ||
| + | <center> | ||
| + | <figure> | ||
| + | <img src="https://savchenkosolutions.com/3/3.1.5/statement.png" | ||
| + | loading="lazy" width="130" /> | ||
| + | <figcaption> | ||
| + | For problem $3.1.5$ | ||
| + | </figcaption> | ||
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| + | </center> | ||
| + | <p> | ||
| + | </p> | ||
| + | |||
| + | <h3>Solution</h3> | ||
| + | <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 | ||
| + | $$ OX:F=-mg\sin\varphi=\fbox{$-mg\frac{x}{l}$} $$ | ||
| + | Considering that the oscillations are harmonic, their amplitude is not large, and the magnitude of the deviation is much less than the length of the thread: $(x\ll l)$. Let us use the expression for small angles $\cos\alpha\approx1-\frac{\alpha^2}{2};\quad\sin\alpha\approx\alpha$ | ||
| + | $$ 1-\cos\alpha\approx\frac{x^2}{2l^2} $$ | ||
| + | Changing the position of the ball vertically | ||
| + | $$ \Delta h = l \cdot (1-\cos\alpha)=\frac{x^2}{2l} $$ | ||
| + | Where does the change in potential energy come from? | ||
| + | $$ \Delta U = mg\cdot\Delta h = \boxed{mg\frac{x^2}{2l}} $$ | ||
| + | </p> | ||
| + | <p style="text-align: right; font-style: italic; font-size: 14;"> | ||
| + | Aliaksandr Melnichenka | ||
| + | </p> | ||
| + | <h3>Alternative solution</h3> | ||
| + | <p> | ||
| + | Alternatively, one could write the change in energy as an integral of the work done on the interval $dx$: | ||
| + | $$ U=\int_{0}^{U}dU=\int_{0}^{x}mg\frac{x}{l}dx=\boxed{\frac{mgx^2}{2l}} $$ | ||
| + | </p> | ||
| + | <p style="text-align: right; font-style: italic; font-size: 14;"> | ||
| + | Dzikan Mikita<br> | ||
| + | </p> | ||
| + | |||
| + | <h4>Answer</h4> | ||
| + | <p> | ||
| + | $$F=-mg\frac{x}{l};\quad U=\frac{mgx^2}{2l}$$ | ||
| + | </p> | ||
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| <body style=""> | |||
| <header style="text-align:center;"> | |||
| <div id = "logo"> | |||
| <span><img src = "../../img/book.png"><span><span>Savchenko Solutions</span> | |||
| </div> | |||
| <p class="author"> | |||
| Solutions of Savchenko Problems in Physics <br> | |||
| <i><b>knowledge must be free</b></i> | |||
| </p> | |||
| </header> | |||
| <h3 id="back-link"><a href="../../#3.1">$\leftarrow$Back</a></h3> | |||
| <h3> Statement </h3> | |||
| <p> | |||
| $3.1.5.$ 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 $x$? | |||
| </p> | |||
| <center> | |||
| <figure> | |||
| <img src="https://savchenkosolutions.com/3/3.1.5/statement.png" | |||
| loading="lazy" width="130" /> | |||
| <figcaption> | |||
| For problem $3.1.5$ | |||
| </figcaption> | |||
| </figure> | |||
| </center> | |||
| <p> | |||
| </p> | |||
| <h3>Solution</h3> | |||
| <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 | |||
| $$ OX:F=-mg\sin\varphi=\fbox{$-mg\frac{x}{l}$} $$ | |||
| Considering that the oscillations are harmonic, their amplitude is not large, and the magnitude of the deviation is much less than the length of the thread: $(x\ll l)$. Let us use the expression for small angles $\cos\alpha\approx1-\frac{\alpha^2}{2};\quad\sin\alpha\approx\alpha$ | |||
| $$ 1-\cos\alpha\approx\frac{x^2}{2l^2} $$ | |||
| Changing the position of the ball vertically | |||
| $$ \Delta h = l \cdot (1-\cos\alpha)=\frac{x^2}{2l} $$ | |||
| Where does the change in potential energy come from? | |||
| $$ \Delta U = mg\cdot\Delta h = \boxed{mg\frac{x^2}{2l}} $$ | |||
| </p> | |||
| <p style="text-align: right; font-style: italic; font-size: 14;"> | |||
| Aliaksandr Melnichenka | |||
| </p> | |||
| <h3>Alternative solution</h3> | |||
| <p> | |||
| Alternatively, one could write the change in energy as an integral of the work done on the interval $dx$: | |||
| $$ U=\int_{0}^{U}dU=\int_{0}^{x}mg\frac{x}{l}dx=\boxed{\frac{mgx^2}{2l}} $$ | |||
| </p> | |||
| <p style="text-align: right; font-style: italic; font-size: 14;"> | |||
| Dzikan Mikita<br> | |||
| </p> | |||
| <h4>Answer</h4> | |||
| <p> | |||
| $$F=-mg\frac{x}{l};\quad U=\frac{mgx^2}{2l}$$ | |||
| </p> | |||
| <footer class="row container"> | |||
| <br> | |||
| <p> | |||
| <small> © <strong>Savchenko Solutions</strong>, 2023-2024 <br></small> | |||
| </p> | |||
| <p> | |||
| <small>All rights belong to the authors. <br> Commercial use of materials - with the written permission of the authors. <br> alex@savchenkosolutions.com <br></small> | |||
| </p> | |||
| </footer> | |||
| </body> | |||
| </html> | |||