Two particles of masses \(m_{1}\) and \(m_{2}\) are attached to the ends of a light spring. The natural length of the spring is \(l\), and its tension is \(k\) times its extension. Initially, the particles are at rest, with \(m_{1}\) at a height \(l\) above \(m_{2} .\) At \(t=0, m_{1}\) is projected vertically upward with velocity \(v\). Find the positions of the particles at any subsequent time (assuming that \(v\) is not so large that the spring is expanded or compressed beyond its elastic limit).

We begin by setting up a coordinate system. Let's say the initial height of \(m_{1}\) above \(m_{2}\) is the reference height, and we'll call this height "h". Define positive y to be upward in height. So initially, we have \(y_{1} = h\) and \(y_{2} = 0\). Now, let's analyze the forces acting on both particles. For \(m_{1}\): - Gravitational force is acting downwards with a magnitude of \(m_{1}g\). - Tension in the spring is acting upwards with a magnitude of \(k(y_{2} - y_{1} + l)\). For \(m_{2}\): - Gravitational force is acting downwards with a magnitude of \(m_{2}g\). - Tension in the spring is acting upwards with a magnitude of \(k(y_{1} - y_{2} - l)\).

Using Newton's Second Law of Motion for each particle, we have: For \(m_{1}\): $$m_{1} \frac{d^2 y_{1}}{dt^2} = k(y_{2} - y_{1} + l) - m_{1}g$$ For \(m_{2}\): $$m_{2} \frac{d^2 y_{2}}{dt^2} = k(y_{1} - y_{2} - l) - m_{2}g$$

The equations of motion are coupled second-order differential equations. We can solve them using standard methods such as substitution or matrix methods. First, we have the initial conditions: at \(t=0\), \(y_{1}(0) = h\), \(y_{2}(0) = 0\), \(\frac{dy_{1}}{dt}(0) = v\), and \(\frac{dy_{2}}{dt}(0) = 0\). To solve the equations, we can obtain expressions for the second derivatives \(\frac{d^2 y_{1}}{dt^2}\) and \(\frac{d^2 y_{2}}{dt^2}\) as a function of time: $$\frac{d^2 y_{1}}{dt^2}(t) = \frac{k(y_{2}(t) - y_{1}(t) + l)}{m_{1}} - g$$ $$\frac{d^2 y_{2}}{dt^2}(t) = \frac{k(y_{1}(t) - y_{2}(t) - l)}{m_{2}} - g$$ Next, we integrate both expressions twice with respect to time and apply the initial conditions to obtain the positions of particles at any time t: $$y_{1}(t) = h + vt + \int_{0}^{t} \int_{0}^{t^*} \left(\frac{k(y_{2}(t^{**}) - y_{1}(t^{**}) + l)}{m_{1}} - g\right) dt^{**} dt^*$$ $$y_{2}(t) = \int_{0}^{t} \int_{0}^{t^*} \left(\frac{k(y_{1}(t^{**}) - y_{2}(t^{**}) - l)}{m_{2}} - g\right) dt^{**} dt^*$$ These expressions represent the position of particles \(m_{1}\) and \(m_{2}\) at any given time "t" with the given initial conditions.