Question

What is the energy of the spring-mass system when the mass first passes through the equilibrium position? (you may wish to include a logical test to help you find when this occurs)?

Answer 1

The spring-mass system moves by simple harmonic motion, where there is a continuous conversion from elastic potential energy to kinetic energy and viceversa.

The total mechanical energy of the system at any moment of the motion is
$E=U+K= \frac{1}{2}kx^2 + \frac{1}{2}mv^2$
where the first term U is the elastic potential energy, with k being the spring constant and x the displacement of the spring with respect to its rest position, and the second term K is the kinetic energy, with m being the mass of the object attached to the spring and v its speed.

The total energy E is constant during the oscillation of the spring, but the values of U and K change. In fact, when the displacement of the spring is maximum (x is maximum), then all the energy is potential energy U, because the speed of the object is zero (it's the moment when the mass is changing direction). On the contrary, when the mass crosses the equilibrium position (rest position) of the spring, then the potential energy is zero (U=0) because the displacement is zero (x=0), and so all the energy is kinetic energy of the motion, and so K is maximum.