Calculate the ratio of the kinetic energy to the potential energy of a simple harmonic oscillator when its displacement is 1/8 of its amplitude. (The answer is an integer.)
Approach: Choose a specific trigonometric form for the position function x(t). It doesn't matter which since the answer doesn't depend on the initial conditions. Let the amplitude be 'A' (which will cancel when you compute the ratio). Determine v(t) that corresponds to your choice of x(t). Apply the given displacement condition to determine what your choice of x(t) was equal to (you don't need to try and compute a time: the result holds for any simple harmonic oscillator). From this directly compute the desired ratio. The trig identity sin2(è)+cos2(è)=1, which as discussed in class is equivalent to the conservation of energy here, will definitely be helpful.
Physics - drwls, Thursday, November 15, 2012 at 5:58am
P.E. is proportional to the square of displacement. When displacement is 1/8 of max, the P.E. is 1/64 of the maximum value, which is also the total energy. The rest (63/64 of total energy) is kinetic energy.
The KE/PE ratio at the time in question is therefore 63.
Physics - Alphonse, Wednesday, November 21, 2012 at 7:21pm
Thanks for the help!