Spring Motion: Difference between revisions
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==Spring Motion== | ==Spring Motion== | ||
While the momentum you have dealt with before may have followed a linear path based on a constant force, spring motion is involved with a changing net force. As you may have noticed when trying to compress or stretch a helical spring, it resists motion (with high opposing force) most when its length is farthest from the relaxed length in either direction. | |||
This force alone would indicate a movement along the length and direction of the spring, oscillating based on the addition of a mass. | |||
===A Mathematical Model=== | ===A Mathematical Model=== | ||
The Force due to a spring follows the relationship <math> F_{spring} = -k_s*s*L_{hat} </math> | |||
where <math> k_s </math> is the specific stiffness of the spring, <math> s </math> is the change in length of the spring from the current to the relaxed length <math> s= L - L_0 </math> and <math> L_{hat} </math> is the unit vector indicating the direction of the vector pointing from the spring's origin to the mass. | |||
Spring motion can be predicted using the momentum principle <math> dp=F_{net}*dt </math> and the position-update formula <math> dr= p_f/m*dt </math> Because spring motion involves oscillation, it is important to iterate with a small enough change in time to capture the full path of the spring-mass system. | |||
===A Computational Model=== | ===A Computational Model=== | ||
Follow the link to check out a 3D model of Spring Motion. The code displays a spring with one fixed end on a ceiling and a mass attached to the free end. | |||
https://trinket.io/glowscript/3587ff8cbe | |||
The pink arrow represents the magnitude and direction of the mass's momentum. | |||
The yellow arrow represents the changing net force, Force of gravity + Force of the spring, on the mass (scaled down to 10%). | |||
Examine the orange curve after the simulation ends to follow the path of the mass's motion. | |||
Try an initial mass location to give a straight vertical path! | |||
=Simple Example= | =Simple Example= | ||
Revision as of 02:32, 1 December 2015
by Caroline Ware
Spring Motion
While the momentum you have dealt with before may have followed a linear path based on a constant force, spring motion is involved with a changing net force. As you may have noticed when trying to compress or stretch a helical spring, it resists motion (with high opposing force) most when its length is farthest from the relaxed length in either direction. This force alone would indicate a movement along the length and direction of the spring, oscillating based on the addition of a mass.
A Mathematical Model
The Force due to a spring follows the relationship [math]\displaystyle{ F_{spring} = -k_s*s*L_{hat} }[/math] where [math]\displaystyle{ k_s }[/math] is the specific stiffness of the spring, [math]\displaystyle{ s }[/math] is the change in length of the spring from the current to the relaxed length [math]\displaystyle{ s= L - L_0 }[/math] and [math]\displaystyle{ L_{hat} }[/math] is the unit vector indicating the direction of the vector pointing from the spring's origin to the mass.
Spring motion can be predicted using the momentum principle [math]\displaystyle{ dp=F_{net}*dt }[/math] and the position-update formula [math]\displaystyle{ dr= p_f/m*dt }[/math] Because spring motion involves oscillation, it is important to iterate with a small enough change in time to capture the full path of the spring-mass system.
A Computational Model
Follow the link to check out a 3D model of Spring Motion. The code displays a spring with one fixed end on a ceiling and a mass attached to the free end. https://trinket.io/glowscript/3587ff8cbe The pink arrow represents the magnitude and direction of the mass's momentum. The yellow arrow represents the changing net force, Force of gravity + Force of the spring, on the mass (scaled down to 10%). Examine the orange curve after the simulation ends to follow the path of the mass's motion. Try an initial mass location to give a straight vertical path!
Simple Example
In this example, the electric field is equal to [math]\displaystyle{ E = \left(E_x, 0, 0\right) }[/math]. The initial location is A and the final location is C. In order to find the potential difference between A and C, we use [math]\displaystyle{ dV = V_C - V_A }[/math].
Since there are no y and z components of the electric field, the potential difference is [math]\displaystyle{ dV = -\left(E_x*\left(x_1 - 0\right) + 0*\left(-y_1 - 0\right) + 0*0\right) = -E_x*x_1 }[/math]
Let's say there is a location B at [math]\displaystyle{ \left(x_1, 0, 0\right) }[/math]. Now in order to find the potential difference between A and C, we need to find the potential difference between A and B and then between B and C.
The potential difference between A and B is [math]\displaystyle{ dV = V_B - V_A = -\left(E_x*\left(x_1 - 0\right) + 0*0 + 0*0\right) = -E_x*x_1 }[/math].
The potential difference between B and C is [math]\displaystyle{ dV = V_C - V_B = -\left(E_x*0 + 0*\left(-y_1 - 0\right) + 0*0\right) = 0 }[/math].
Therefore, the potential difference A and C is [math]\displaystyle{ V_C - V_A = \left(V_C - V_B\right) + \left(V_B - V_A\right) = E_x*x_1 }[/math], which is the same answer that we got when we did not use location B.
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