1) 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?
See attached file for full problem description.© BrainMass Inc. brainmass.com March 4, 2021, 7:31 pm ad1c9bdddf
Hello, solution is here. I have attached a file. Contends are copied and pasted below as well. However, the graph will appear only in the attached document.
1) 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.
For systems that obey Hooke's law, the extension is proportional to the applied force.
F = -kx, where x is the distance the spring is elongated by, F is the restoring force exerted by the spring, and k is the spring constant or force constant of the spring.
In your problem, extension is (L1 - Lo)
F = - k (L1 - Lo)
Since the system is at equilibrium, we can say that the gravitational force balances the spring force. i.e. net force = 0
Gravitational force + spring force =0 ...
Solution is provided in a 3-page word document along with a graph for Morse function. Every step is clearly written in finding the solution to these two problems.