EJSS simple harmonic motion vertical spring mass model with sensor

based on models and ideas by

- lookang http://weelookang.blogspot.sg/2014/02/ejss-shm-model-with-vs-x-and-v-vs-x.html
- lookang http://weelookang.blogspot.sg/2010/06/ejs-open-source-simple-harmonic-motion.html?q=SHM
- lookang http://weelookang.blogspot.sg/2013/02/ejs-open-source-vertical-spring-mass.html?q=vertical+spring
- Wolfgang Christian and Francisco Esquembre http://www.opensourcephysics.org/items/detail.cfm?ID=13103

## 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:

$ 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).

where the terms

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

In equilibrium position,

where $ y_{0} is the new position or length extra extended beyond the natural length

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 $.

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 hereIn 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

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

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

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 $