A massless spring has unstretched length lo and force constant k. One end is now attached to the ceiling and a mass m is hung from the other. The equilibrium length of the spring is now l1.

(a) Write down the condition that determines l1. Suppose now that the spring is stretched a further   distance x beyond its new equilibrium length. Show that the net force (spring plus gravity) on the mass is F = -kx. That is, the net force obeys Hooke's Law, when x is the distance from the equilibrium position - a very useful result, which lets us treat a mass on a vertical spring just as if it were horizontal.

(b) Prove the same result by showing that the net potential energy (spring plus gravity) has the form U(x) = const + (1/2)kx2.

2) The potential energy of two atoms in a molecule can sometimes be approximated by the Morse function, U(r) = A [(e^{(R - r)/S} - 1)² - 1] where r is the distance between the two atoms A, R, and S are positive constants with S <<< R. Sketch this function for 0 < r < infinity. Find the equilibrium separation ro, at which U(r) is minimum. Now write r = ro + x so that x is the displacement from equilibrium, and show that, for small displacements, U has the approximate form U = const + (1/2)kx² . That is Hooke's Law applies. What is the force constant k?