Problem 5. Vertical oscillations of a load of mass $m$ on a spring with stiffness $k$ in a viscous medium are described by the equation $x(t)=a e^{-2 t}+b e^{-t}+m g / k$, where $a, b$ are constants depending on the initial conditions, $t$ is time. At the initial moment, the load was taken out of equilibrium. Given $\mathrm{mg} / \mathrm{k}=1$ and $b=1-2 a$, find the values of the constant $a$ for which the spring will be in an unstretched state ( $x=0$ ) twice during the motion.
Solution. The substitution $y=e^{-t}$ reduces the problem to the following: find the values of the parameter $a$ for which the equation $a y^{2}+(1-2 a) y+1=0$ has two roots in the interval $(0 ; 1)$.

The conditions for this are:

$$
D=4 a^{2}-8 a+1 \geq 0 ; a f(0)=a>0 ; \text { af }(1)=a(a+(1-2 a)+1)>0 ; y_{0}=\frac{2 a-1}{2 a} \in(0 ; 1)
$$

From this, the answer is: $a \in\left(1+\frac{\sqrt{3}}{2} ; 2\right)$.

Answer: $a \in\left(1+\frac{\sqrt{3}}{2} ; 2\right)$.

Answer to variant 212: $a \in(4+2 \sqrt{3} ;+\infty)$.