# The Velocity of a Classic SHO

summary:

The scalar term is equal to 1 for every velocity because the problem is classical.

description:

When one take the time derivative of the oscillator, $\frac{d t}{d t} = 1$. I puzzled over that obvious result for days, since its meaning was not clear. Then I recalled in relativistic physics, one takes the derivative with respect to the interval tau, so $\frac{d t}{d \tau} = \gamma$. For tiny values of velocity, gamma will equal 1. What people often graph is the velocity parameterized by time. They don't wish to treat the 4-velocity like a 4-velocity. If you keep the books consistent, you can get fun surprises.

command:
q_graph -out sho-x_and_v -dir sho -box 10 -command 't_function -mqs 2 -mqs 1 -t_function polynomial -t_function one -x_function cos -x_function polynomial -y_function zero -y_function zero -z_function zero -z_function zero 1 2 1 2 1 3 0 0 0 1 0 0 -shift "-11 5 0 0" -pi 8 -n_steps 1600' -color yellow -command 't_function -mqs 2 -mqs 1 -t_function one -t_function one -x_function sin -x_function polynomial -y_function zero -y_function zero -z_function zero -z_function zero -shift "0 5 0 0" -pi 8 -n_steps 1600 -- 1 2 1 2 1 3 0 0 0 -2 0 0' -color blue
math
equation:

$(t, x, y, z) = (t, cos( 2 t + 3), 0, 0)$
$(\gamma, \gamma \beta_x, \gamma \beta_y, \gamma \beta_z) = (1, -2 sin( 2 t + 3), 0, 0)$

tags
Physics Tag:
simple harmonic oscillator
Math Tag:
sine
cosine
Programming Tag:
command line quaternions
t_function