Two identical bodies of mass \(m\) are connected by a spring of stiffiness \(k\) and constrained to move in rectilinear motion as shown. Derive the equations of motion for the system.

Newton's second law of motion states that the force acting on an object is equal to its mass times its acceleration: \(F = ma\). Hooke's law states that the force exerted by a spring is proportional to its displacement from the equilibrium position: \(F = -kx\). In this problem, we need to find the equations of motion for the two bodies. Let \(x_1\) and \(x_2\) denote the displacements of mass 1 and mass 2 from their respective equilibrium positions, such that the positive direction is to the right.

For mass 1, there are two forces acting on it: the force due to the spring connecting it to mass 2, and the force due to its own inertia. According to Hooke's law, the force exerted by the spring is \(-k(x_1 - x_2)\), where the negative sign indicates that the force is always directed toward the equilibrium position. The force due to its own inertia is simply \(-mx_1''\). For mass 2, there are two forces acting on it: the force due to the spring connecting it to mass 1, and the force due to its own inertia. According to Hooke's law, the force exerted by the spring is \(-k(x_2 - x_1)\), where the negative sign indicates that the force is always directed toward the equilibrium position. The force due to its own inertia is simply \(-mx_2''\).

Now, we will apply Newton's second law for each mass: For mass 1: \(-k(x_1 - x_2) = m x_1''\) For mass 2: \(-k(x_2 - x_1) = m x_2''\)

We can rewrite the equations from Step 3 as follows: Mass 1: \(m x_1'' + k(x_1 - x_2) = 0\) Mass 2: \(m x_2'' + k(x_2 - x_1) = 0\)

These two equations represent the equations of motion for the system of two identical masses connected by a spring: \(m x_1'' + k(x_1 - x_2) = 0\) \(m x_2'' + k(x_2 - x_1) = 0\) These are the differential equations that describe the motion of the two masses in the given system.