## Monday, March 3, 2014

### EJSS vertical spring model with sensor

EJSS SHM vertical spring mass model with sensor
EJSS simple harmonic motion vertical spring mass model with sensor
based on models and ideas by

 http://weelookang.blogspot.com/2014/03/ejss-vertical-spring-model-with-sensor.html EJSS SHM vertical spring mass model with sensor with equilibrium as y=0 https://dl.dropboxusercontent.com/u/44365627/lookangEJSworkspace/export/ejss_model_SHMxvavertical01/SHMxvavertical01_Simulation.html source: https://dl.dropboxusercontent.com/u/44365627/lookangEJSworkspace/export/ejss_src_SHMxvavertical01.zip author: lookang author of EJSS 5.0 Francisco Esquembre

 http://weelookang.blogspot.com/2014/03/ejss-vertical-spring-model-with-sensor.html EJSS SHM vertical spring mass model with sensor with sensor as y =0 https://dl.dropboxusercontent.com/u/44365627/lookangEJSworkspace/export/ejss_model_SHMxvavertical01/SHMxvavertical01_Simulation.html source: https://dl.dropboxusercontent.com/u/44365627/lookangEJSworkspace/export/ejss_src_SHMxvavertical01.zip author: lookang author of EJSS 5.0 Francisco Esquembre

## Assumption:

Motion approximates SHM when the spring does not exceed limit of proportionality during oscillations.

 http://weelookang.blogspot.com/2014/02/ejss-vertical-spring-mass-model.html EJSS SHM vertical spring mass model with y vs t, v vs t and a vs t graph suitable for understanding lowering equilibrium position effects of mass m https://dl.dropboxusercontent.com/u/44365627/lookangEJSworkspace/export/ejss_model_SHMxvavertical/SHMxvavertical_Simulation.html source: https://dl.dropboxusercontent.com/u/44365627/lookangEJSworkspace/export/ejss_src_SHMxvavertical.zip author: lookang author of EJSS 5.0 Francisco Esquembre

## The equations that model the motion of the vertical spring mass system are:

Mathematically, the restoring force $F$ is given by

$F = - k y$

where $F$  is the restoring elastic force exerted by the spring (in SI units: N), $k$ is the spring constant (N·m−1), and $y$ is the displacement from the equilibrium position (in m).

Thus, this model is simplified by http://physics.ucsc.edu/~josh/6A/book/harmonic/node13.html assumes

$\frac{\delta y}{\delta t} = v_{y}$

$\frac{\delta v_{y}}{\delta t} = -\frac{k}{m}(y)$

where the terms

$-\frac{k}{m}(y)$ represents the restoring force component as a result of the spring extending and compressing.

## Why is the equation $-ky = ma$ ?

josh explains it well here

In equilibrium position,
$-k y_{e} -mg =0$

where $y_{e} is the position or length extended beyond the natural length$ y_{e} = -\frac{mg}{k} $using$ F_{net} = ma  -k y_{0} -mg = ma $where$ y_{0} is the  new position or length extra extended beyond the natural length

$-k y_{0} - (- k y_{e}) = ma$

$-k (y_{0} - y_{e}) = ma$

renaming $(y_{0} - y_{e})$ as $y$

we get

$-k y = ma$

which keeps the equilibrium constantly at zero and the mass $m$ and spring constant $k$ effects the angular frequency $\omega = \sqrt{ (\frac{k}{m})}$. So gravity has no effect on the oscillation frequency $\omega$.

## Calculations used in the model:

Equilibrium height or position $h$: typically it is zero but it may be displaced with a different origin thus using this equation helps

$h = y_{sensor}$

Amplitude $x_{0}$ is defined as magnitude of the maximum displacement from the
equilibrium position. Since the motion starts when zero initial velocity, it is generally true that

$x_{0} = y$ when  $t = 0$

Period $T$ Time taken for one complete oscillation is easily to determined visually but it can be a challenge to pre-determined even before the model runs $t = 0$. The way used in the model is to determine period by assuming

$T = 2 \pi \sqrt { \frac{m}{k}}$

Frequency $f$ Number of oscillations performed per unit time. Mathematically the period is related to frequency as a reciprocal of the other.

$f = \frac{1}{T}$

Angular Frequency $\omega$ is the about of angle in radian covered per unit time. Thus, if knowing $T$ is the time for one complete oscillation which is $2 \pi$ radians, thus

$\omega = \frac{2 \pi}{T} = 2 \pi f$