Tuesday, February 18, 2014

EJSS vertical spring mass model

EJSS SHM vertical spring mass model with y vs t, v vs t and a vs t graph
EJSS simple harmonic motion vertical spring mass model with y vs t, v vs t and a vs t graph
based on models and ideas by

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

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
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 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 assumes 

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


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

where the terms

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

$ - g $ represents the gravity force component as a result of Earth's pull.


What is SHM?

Simple harmonic motion is typified by the motion of a mass on a spring when it is subject to the linear elastic restoring force given by Hooke's Law. The motion is sinusoidal in time and demonstrates a single resonant frequency. In order for simple harmonic motion to take place, the net force of the object at the end of the pendulum must be proportional to the displacement. In other words, oscillations are periodic variations in the value of a physical quantity about a central or equilibrium value.

Once the mass is displaced from its equilibrium position, it experiences a net restoring force. As a result, it accelerates and starts going back to the equilibrium position. When the mass moves closer to the equilibrium position, the restoring force decreases. At the equilibrium position, the net restoring force vanishes. However, at y = 0, the mass has momentum because of the impulse that the restoring force has imparted. Therefore, the mass continues past the equilibrium position, compressing the spring. A net restoring force then tends to slow it down, until its velocity reaches zero, whereby it will attempt to reach equilibrium position again.


If motion starts at the equilibrium position and starts to move to the positive direction solutions to the defining equation are:

not possible, as it is in equilibrium.


If the motion starts to the negative amplitude position:

$ y = - y_{o} cos ( \omega t ) = y_{o} sin ( \omega t -  \frac{\pi }{2} )$
$ y = - y_{o} cos ( \omega t ) = y_{o} sin ( \omega t -  \frac{\pi }{2} )$


$ v = y_{o} \omega sin ( \omega t ) = y_{o} \omega cos ( \omega t -  \frac{\pi }{2} )$
$ v = y_{o} \omega sin ( \omega t ) = y_{o} \omega cos ( \omega t -  \frac{\pi }{2} )$


$ a =  y_{o} \omega^{2} cos ( \omega t ) = - y_{o} \omega^{2} sin ( \omega t -  \frac{\pi }{2} )$
$ a =  y_{o} \omega^{2} cos ( \omega t ) = - y_{o} \omega^{2} sin ( \omega t -  \frac{\pi }{2} )$




therefore , in general:

$ y = y_{o} sin ( \omega t - \phi ) $

where

$ \phi = \pi/2 $ for a starting position of  $ y = - y_{o}$
$ \phi = - \pi/2 $ for a starting position of  $ y = y_{o}$


$ v = y_{o} \omega cos ( \omega t - \phi ) $

$ a = - y_{o} \omega^{2} sin ( \omega t - \phi ) $