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| <title>The body of mass m is connected by two springs of stiffness k_1 and k_2 with fixed walls, the springs are not initially deformed. When vibrations occur, the greatest acceleration of the body is a. Find the maximum deviation of the body from the equilibrium position and the maximum forces with which the springs act on the walls.</title> | | <title>The body of mass m is connected by two springs of stiffness k_1 and k_2 with fixed walls, the springs are not initially deformed. When vibrations occur, the greatest acceleration of the body is a. Find the maximum deviation of the body from the equilibrium position and the maximum forces with which the springs act on the walls.</title> |
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| <h3 id="back-link"><a href="../../#2.1">$\leftarrow$Back</a></h3> | | <h3 id="back-link"><a href="../../#2.1">$\leftarrow$Back</a></h3> |
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| <h3> Statement </h3> | | <h3> Statement </h3> |
| <p> | | <p> |
| $2.1.16.$ The body of mass $m$ is attached to two consecutive springs of stiffness $k_1$ and $k_2$. A constant force $F$ is applied to the free end of the spring chain. What is the total elongation of the springs, if the vibrations have already stopped? | | $2.1.16.$ The body of mass $m$ is attached to two consecutive springs of stiffness $k_1$ and $k_2$. A constant force $F$ is applied to the free end of the spring chain. What is the total elongation of the springs, if the vibrations have already stopped? |
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| For problem $2.1.16$ | | For problem $2.1.16$ |
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| <h3>Solution</h3> | | <h3>Solution</h3> |
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| loading="lazy" width="350" /> | | loading="lazy" width="350" /> |
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| Series spring system | | Series spring system |
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| <p> | | <p> |
| 1. When springs are connected in series, their deformation will be different with the same acting force, this circumstance allows us to determine the overall stiffness of the springs as follows | | 1. When springs are connected in series, their deformation will be different with the same acting force, this circumstance allows us to determine the overall stiffness of the springs as follows |
| $$ \Delta x_0 = \Delta x_1 + \Delta x_2 = \frac{F}{k_1}+\frac{F}{k_2} = \frac{F}{k_0} $$ | | $$ \Delta x_0 = \Delta x_1 + \Delta x_2 = \frac{F}{k_1}+\frac{F}{k_2} = \frac{F}{k_0} $$ |
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| $$ \boxed{k_0 = \frac{k_1 k_2}{k_1 + k_2}} $$ | | $$ \boxed{k_0 = \frac{k_1 k_2}{k_1 + k_2}} $$ |
| 2. The combined effect on the mass of the springs when the mass is at rest will be equal to the applied force $F$ | | 2. The combined effect on the mass of the springs when the mass is at rest will be equal to the applied force $F$ |
| $$ k_0 \Delta x_0 = F $$ | | $$ k_0 \Delta x_0 = F $$ |
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| $$ \boxed{ \Delta x_0 = \frac{F(k_1 + k_2)}{k_1 k_2}} $$ | | $$ \boxed{ \Delta x_0 = \frac{F(k_1 + k_2)}{k_1 k_2}} $$ |
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| <h3> | | <h3> |
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| <h4>Answer</h4> | | <h4>Answer</h4> |
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| $$x = F(k_1 + k_2)/(k_1k_2)$$ | | $$x = F(k_1 + k_2)/(k_1k_2)$$ |
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