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 $
No comments:
Post a Comment