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| <meta name="author" content="Aliaksandr Melnichenka"> | | <meta name="author" content="Aliaksandr Melnichenka"> |
| <meta name="date" content="2023-10" scheme="YYYY-MM"> | | <meta name="date" content="2023-10" scheme="YYYY-MM"> |
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| <h2>Solutions of Savchenko Problems in Physics</h2> | | <h2>Solutions of Savchenko Problems in Physics</h2> |
| <p class="author"> | | <p class="author"> |
| Aliaksandr Melnichenka <br/> | | Aliaksandr Melnichenka <br/> |
| October 2023 | | October 2023 |
| </p> | | </p> |
| </header> | | </header> |
| | | |
| <h3 id="back-link"><a href="../../#2.1">$\leftarrow$Back</a></h3> | | <h3 id="back-link"><a href="../../#2.1">$\leftarrow$Back</a></h3> |
| <img src="https://savchenkosolutions.com/2/2.1.15/sol2.png" | | <img src="https://savchenkosolutions.com/2/2.1.15/sol2.png" |
| loading="lazy" width="250" /> | | loading="lazy" width="250" /> |
| <figcaption> | | <figcaption> |
| For problem $2.1.15$ | | For problem $2.1.15$ |
| </figcaption> | | </figcaption> |
| </figure> | | </figure> |
| </center> | | </center> |
| <p> | | <p> |
| </p> | | </p> |
| | | |
| <h3>Solution</h3> | | <h3>Solution</h3> |
| <p> | | <p> |
| <center> | | <center> |
| <figure> | | <figure> |
| <img src="https://savchenkosolutions.com/2/2.1.15/sol1.png" | | <img src="https://savchenkosolutions.com/2/2.1.15/sol1.png" |
| loading="lazy" width="230" /> | | loading="lazy" width="230" /> |
| <figcaption> | | <figcaption> |
| Equivalent system of springs | | Equivalent system of springs |
| </figcaption> | | </figcaption> |
| </figure> | | </figure> |
| </center> | | </center> |
| | | |
| <p> | | <p> |
| 1. The springs in this problem are connected in parallel, their deformation is the same | | 1. The springs in this problem are connected in parallel, their deformation is the same |
| $$ Δx_1 = Δx_2 = Δx $$ | | $$ Δx_1 = Δx_2 = Δx $$ |
| 2. The force acting on the mass from the springs is determined as the sum | | 2. The force acting on the mass from the springs is determined as the sum |
| $$ F = F + F_1 $$ | | $$ F = F + F_1 $$ |
| alternatively | | alternatively |
| $$ k \Delta x = k_1 \Delta x + k_2 \Delta x $$ | | $$ k \Delta x = k_1 \Delta x + k_2 \Delta x $$ |
| 3. Let us write down the equation of motion of the mass under the action of an equivalent spring with stiffness ko, which will allow us to determine the maximum displacement | | 3. Let us write down the equation of motion of the mass under the action of an equivalent spring with stiffness ko, which will allow us to determine the maximum displacement |
| $$ ma = (k_1 + k_2) \Delta x_{max} $$ | | $$ ma = (k_1 + k_2) \Delta x_{max} $$ |
| | | |
| $$ \Delta x_{max} = ma/(k_1 + k_2) $$ | | $$ \Delta x_{max} = ma/(k_1 + k_2) $$ |
| 4. Maximum values of forces acting on the mass | | 4. Maximum values of forces acting on the mass |
| $$ {F_1}_{max} = k_1 \Delta x_{max} $$ | | $$ {F_1}_{max} = k_1 \Delta x_{max} $$ |
| | | |
| $$ {F_2}_{max} = k_2 \Delta x_{max} $$ | | $$ {F_2}_{max} = k_2 \Delta x_{max} $$ |
| <p> | | <p> |
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| </p> | | </p> |
| </p> | | </p> |
| | | |
| <h4>Answer</h4> | | <h4>Answer</h4> |
| <p> | | <p> |
| <p> | | <p> |
| $$x_{max} = ma/(k_1 + k_2);$$ | | $$x_{max} = ma/(k_1 + k_2);$$ |
| $${F_1}_{max} = k_1x_{max}$$ | | $${F_1}_{max} = k_1x_{max}$$ |
| $${F_2}_{max} = k_2x_{max}$$ | | $${F_2}_{max} = k_2x_{max}$$ |
| </p> | | </p> |
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