Translated 3.1.1-3.1.17
en/3.1.16.md
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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> | ||
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| + | <h3 id="back-link"><a href="../../#3.1">$\leftarrow$Back</a></h3> | ||
| + | |||
| + | <h3> Statement </h3> | ||
| + | <p> | ||
| + | $3.1.16.$ Determine within what limits the tension force of a mathematical pendulum is if the amplitude of oscillations $x_0$ is much smaller than the length of the string $l$, the mass of the pendulum is $m$. | ||
| + | </p> | ||
| + | |||
| + | <h3>Solution</h3> | ||
| + | <p> | ||
| + | We will find the tension force of the thread as | ||
| + | $$ T_\text{min}=mg\cos\varphi $$ | ||
| + | Let's use the approximation, for $\varphi\ll1$, $\cos\varphi=1-\frac{\varphi^2}{2}\quad(1)$: | ||
| + | $$ mg\cos\varphi \approx mg\left(1-\frac{\varphi^2}{2}\right) $$ | ||
| + | |||
| + | $$ \boxed{T_\text{min} =mg\left(1-\frac{x_0^2}{2l^2}\right)} $$ | ||
| + | Let's write down Newton's second law | ||
| + | $$ ma=T_\text{max}-mg $$ | ||
| + | |||
| + | $$ T_\text{max}=m\left(\frac{\upsilon^2}{l}+g\right) $$ | ||
| + | Also, let's write down the law of conservation of energy | ||
| + | $$ mgl(1-\cos\varphi)=\frac{m\upsilon^2}{2} $$ | ||
| + | |||
| + | $$ \upsilon^2=2gl(1-\cos\varphi) $$ | ||
| + | Given the approximation $(1)$ | ||
| + | $$ v^2\approx gl\varphi^2=\frac{gx_0^2}{l} $$ | ||
| + | Let's substitute this expression for speed into the expression for $T_\text{max}$: | ||
| + | $$ \boxed{T_\text{max}=mg\left(1+\frac{x_0^2}{l^2}\right)} $$ | ||
| + | |||
| + | </p> | ||
| + | |||
| + | <h4>Answer</h4> | ||
| + | <p> | ||
| + | $$mg\left(1-\frac{x_0^2}{2l^2}\right) < T < mg\left(1+\frac{x_0^2}{l^2}\right)$$ | ||
| + | </p> | ||
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| </head> | |||
| <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> | |||
| $3.1.16.$ Determine within what limits the tension force of a mathematical pendulum is if the amplitude of oscillations $x_0$ is much smaller than the length of the string $l$, the mass of the pendulum is $m$. | |||
| </p> | |||
| <h3>Solution</h3> | |||
| <p> | |||
| We will find the tension force of the thread as | |||
| $$ T_\text{min}=mg\cos\varphi $$ | |||
| Let's use the approximation, for $\varphi\ll1$, $\cos\varphi=1-\frac{\varphi^2}{2}\quad(1)$: | |||
| $$ mg\cos\varphi \approx mg\left(1-\frac{\varphi^2}{2}\right) $$ | |||
| $$ \boxed{T_\text{min} =mg\left(1-\frac{x_0^2}{2l^2}\right)} $$ | |||
| Let's write down Newton's second law | |||
| $$ ma=T_\text{max}-mg $$ | |||
| $$ T_\text{max}=m\left(\frac{\upsilon^2}{l}+g\right) $$ | |||
| Also, let's write down the law of conservation of energy | |||
| $$ mgl(1-\cos\varphi)=\frac{m\upsilon^2}{2} $$ | |||
| $$ \upsilon^2=2gl(1-\cos\varphi) $$ | |||
| Given the approximation $(1)$ | |||
| $$ v^2\approx gl\varphi^2=\frac{gx_0^2}{l} $$ | |||
| Let's substitute this expression for speed into the expression for $T_\text{max}$: | |||
| $$ \boxed{T_\text{max}=mg\left(1+\frac{x_0^2}{l^2}\right)} $$ | |||
| </p> | |||
| <h4>Answer</h4> | |||
| <p> | |||
| $$mg\left(1-\frac{x_0^2}{2l^2}\right) < T < mg\left(1+\frac{x_0^2}{l^2}\right)$$ | |||
| </p> | |||
| <p style="text-align: right; font-style: italic; font-size: 14;"> | |||
| Dzikan Mikita<br> | |||
| </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> | |||
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