Two weightless springs have lengths $l_{1}, l_{2}$ and stiffnesses $k_{1}, k_{2}$. One spring is inserted into the other spring. The ends of the springs are connected in pairs. The other points of the springs do not touch each other. What is the stiffness $k$ of the resulting spring?
Decision:
For certainty, we assume that $l_{1}>l_{2}$. The length $l$ of the composite spring satisfies the inequality $l_{2} > l > l_{1}$.
Suppose that under the action of a force $\bar{F}$ directed along the axis, the spring system is stretched (compressed) by some value $\Delta x$. According to Hooke's law.
$\bar{F} = k \Delta x$ (1).
and the first and second springs are subjected to some forces $\bar{F}$ some forces $\bar{F_{1}}, \bar{F_{2}}$ such that
$\bar{F} = \bar{F_{1}}+\bar{F_{2}}$. (2)
Let $\Delta x > 1_{2} - l_{1}$ for certainty. Then both springs are stretched and the forces $\bar{F_{1}}, \bar{F_{2}}$ are directed in the same direction. Let's project the vector equality to the direction of the axis of the stretched springs:
$F= F_{1} + F_{2}$. (3)
Under the action of forces $\bar{F_{1}}, \bar{F_{2}}$ the springs are stretched relative to their equilibrium positions by the values $l-l_{1} + \Delta x$ and $l_{2} - l + \Delta x$, respectively. By Hooke's law.
$F_{1} = k_{1}(l-l_{1}+ \Delta x)$, (4)
$F_{2} = k_{2}(l_{2}-l+ \Delta x)$. (5)
From Equations (1) through (4), we obtain Eqs.
$k \Delta x = k_{1}(l - l_{1}) + k_{2}(l_{2}-l) + (k_{1}+k_{2}) \Delta x$, (6)
which must be satisfied for any values of $\Delta x$, including and at $\Delta x = 0$. This is possible only if
$k_{1}(l-l_{1})+k_{2}(l_{2}-l)=0$ (7)
Given (6), from (5) we find: $k = k_{1} + k_{2}$.
Decision:
For certainty, we assume that $l_{1}>l_{2}$. The length $l$ of the composite spring satisfies the inequality $l_{2} > l > l_{1}$.
Suppose that under the action of a force $\bar{F}$ directed along the axis, the spring system is stretched (compressed) by some value $\Delta x$. According to Hooke's law.
$\bar{F} = k \Delta x$ (1).
and the first and second springs are subjected to some forces $\bar{F}$ some forces $\bar{F_{1}}, \bar{F_{2}}$ such that
$\bar{F} = \bar{F_{1}}+\bar{F_{2}}$. (2)
Let $\Delta x > 1_{2} - l_{1}$ for certainty. Then both springs are stretched and the forces $\bar{F_{1}}, \bar{F_{2}}$ are directed in the same direction. Let's project the vector equality to the direction of the axis of the stretched springs:
$F= F_{1} + F_{2}$. (3)
Under the action of forces $\bar{F_{1}}, \bar{F_{2}}$ the springs are stretched relative to their equilibrium positions by the values $l-l_{1} + \Delta x$ and $l_{2} - l + \Delta x$, respectively. By Hooke's law.
$F_{1} = k_{1}(l-l_{1}+ \Delta x)$, (4)
$F_{2} = k_{2}(l_{2}-l+ \Delta x)$. (5)
From Equations (1) through (4), we obtain Eqs.
$k \Delta x = k_{1}(l - l_{1}) + k_{2}(l_{2}-l) + (k_{1}+k_{2}) \Delta x$, (6)
which must be satisfied for any values of $\Delta x$, including and at $\Delta x = 0$. This is possible only if
$k_{1}(l-l_{1})+k_{2}(l_{2}-l)=0$ (7)
Given (6), from (5) we find: $k = k_{1} + k_{2}$.