Pencil and paper ready. An ideal mass on a spring is pulled back to its maximum displacement and has 100 joules of potential energy before it is released. For 10 points each:
[10m] What is the potential energy of the system when the mass is at one-half of its maximum displacement? You have 10 seconds.
ANSWER: 25 joules (This is solved by using the fact that U0 = ½kx02, so U = ½k(½x)2 = ¼(½kx2) = ¼U0 = 25 joules, where U0 is initial potential energy, U is current potential energy, k is the spring constant, and x0 is the initial displacement.)
[10e] Assuming no frictional losses, what is the kinetic energy of the mass at one-half of its maximum displacement? Recall that the current potential and kinetic energies sum to the initial potential energy of 100 joules.
ANSWER: 75 joules (This is solved by using the conservation of energy, i.e. K + U = U0, so K = U0 - U = 75 joules, where K is current kinetic energy.)
[10h] In the presence of frictional losses such that the mass loses half its amplitude each cycle, this quantity equals approximately 11 percent. A dynamical system undershoots when this quantity is less than one.
ANSWER: damping ratio [prompt on damping or zeta]
<Adam Silverman, Science - Physics> ~20501~ <Editor: David Bass>