$2.1.16.$ 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?
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 $$ \Delta x_0 = \Delta x_1 + \Delta x_2 = \frac{F}{k_1}+\frac{F}{k_2} = \frac{F}{k_0} $$ $$ \boxed{k_0 = \frac{k_1 k_2}{k_1 + k_2}} $$ 2. The combined effect on the mass of the springs when the mass is at rest will be equal to the applied force $F$ $$ k_0 \Delta x_0 = F $$ $$ \boxed{ \Delta x_0 = \frac{F(k_1 + k_2)}{k_1 k_2}} $$
$$x = F(k_1 + k_2)/(k_1k_2)$$