Energy in Simple Harmonic Motion — Question 1

Question 1

Energy ratios locate a particle without fixing its direction

A mass on an ideal spring has amplitude $A=.10\,\mathrm{m}$ and spring constant $k=80\,\mathrm{N/m}$. At an instant its kinetic energy is three times its spring potential energy. Set potential energy to zero at equilibrium.

1. Find total energy, both allowed positions, and the separate kinetic and potential energies.

2. For mass $m=.20\,\mathrm{kg}$, find both velocity signs allowed at each position. How many phase states satisfy the energy ratio?

Question 1 - Solution

Energy ratios locate a particle without fixing its direction

(a) Derivation and calculation

$$E=\tfrac12kA^2=\tfrac12(80)(.10)^2=.400\,\mathrm J.$$$$K=3U,\quad E=K+U=4U\Rightarrow U=.100\,\mathrm J,\ K=.300\,\mathrm J.$$$$\tfrac12kx^2=E/4\Rightarrow\boxed{x=\pm A/2=\pm.0500\,\mathrm m}.$$

(b) Result and interpretation

$$\tfrac12mv^2=.300\Rightarrow\boxed{v=\pm\sqrt{\frac{.600}{.20}}=\pm1.7321\,\mathrm{m/s}}.$$$$(x,v)=(+.0500,\pm1.7321),\quad(-.0500,\pm1.7321).$$Each position is visited once in each direction during a cycle.

Physical check

$K/E=3/4$ and $U/E=1/4$ sum to unity; energy alone does not determine the velocity sign.