Iterative Prediction of Spring-Mass System
claimed by kgiles7
The Main Idea
A simple spring-mass system is a basic illustration of the momentum principle. The principle of conservation of momentum can be repeatedly applied to predict the system's future motion.
A Mathematical Model
The Momentum Principle provides a mathematical basis for the repeated calculations needed to predicts the system's future motion. The most useful form of this equation for predicting future motion is referred to as the momentum update form, and can be derived by rearranging the Momentum Principle as shown below:
[math]\displaystyle{ {\vec{p}_{f} - \vec{p}_{i} = \vec{F}_{net}{Δt}} }[/math]
[math]\displaystyle{ {\vec{p}_{f} = \vec{p}_{i} + \vec{F}_{net}{Δt}} }[/math]
Velocity Update Formula: [math]\displaystyle{ {\vec{v}_{f} = \vec{v}_{i} + \frac{\vec{F}_{net}}{m}}{Δt} }[/math]
Position Update Formula: [math]\displaystyle{ {\vec{r}_{f} = \vec{r}_{i} + \vec{v}_{avg}{Δt}} }[/math]
A Computational Model
How do we visualize or predict using this topic. Consider embedding some vpython code here Teach hands-on with GlowScript
Examples
Simple Example
The simplest example of a spring mass system is one that moves in only one-direction. Consider a massless spring of length 1.0 m with spring constant 40 N/m. If a 1 kg mass is released from rest while the spring is stretched downward to a length of 1.5 m, what is it's position after 0.2 seconds? The mass oscillates vertically, as shown to the right.
Step 1: Set Parameters Before we begin, we must set some parameters that allow the problem to be solved. Since all movement is in the vertical direction, no vector calculation is needed. However, careful attention must be paid to the direction associated with the object's movement and the forces acting on it. For ease in calculations, assume the origin is the point of attachment of the spring to the ceiling above. AS is customary, the positive y-direction will be up, and the negative-y, down. Additionally, the time step used for each iteration must be small enough that we can assume a constant velocity over the interval, but not so large that solving the problem becomes incredibly time-consuming. An appropriately small time step here is approximately 0.1 seconds.
Step 2: Calculate Initial Values Begin by calculating the object's initial momentum and the sum of the forces acting on it. The initial momentum is simply the product of the initial velocity, which is 0 m/s, as the object is released from rest. The forces acting on the mass are the gravitational force exerted by the Earth and the tension exerted by the spring. The net force can be calculated by summing these two forces.
[math]\displaystyle{ {\vec{p}_{i}} = {{m}\cdot\vec{v}_{i}} }[/math]
[math]\displaystyle{ {\vec{p}_{i}} = {{10kg}\cdot{0m/s}} = {0 Ns} }[/math]
[math]\displaystyle{ {\vec{F}_{grav}} = {mg} }[/math]
[math]\displaystyle{ {\vec{F}_{grav}} = {10kg}\cdot{-9.8 m/s/s} = {-9.8 N} }[/math]
[math]\displaystyle{ {\vec{F}_{spring}} = {-kx} }[/math]
[math]\displaystyle{ {\vec{F}_{spring}} = {-40N/m}\cdot{1.0m-1.5m} = {20 N} }[/math]
[math]\displaystyle{ {\vec{F}_{net}} = {\vec{F}_{grav} +{\vec{F}_{spring}}} }[/math]
[math]\displaystyle{ {\vec{F}_{net}} = {-9.8N} +{20 N} = {10.2} }[/math]
Step 3: Update Momentum (Iteration 1)
Using the momentum update formula, calculate the momentum of the mass at the end of the given time step.
[math]\displaystyle{ {\vec{p}_{f} = \vec{p}_{i} + \vec{F}_{net}{Δt}} }[/math]
[math]\displaystyle{ {\vec{p}_{f} = {0Ns} + {10.2N}\cdot{0.1s} = {1.02Ns}} }[/math]
Step 4: Update Velocity (Iteration 1)
[math]\displaystyle{ {\vec{v}_{f} = \vec{v}_{i} + \frac{\vec{F}_{net}}{m}}{Δt} }[/math]
[math]\displaystyle{ {\vec{v}_{f} = {0m/s} + \frac{1.02N}{1 kg}}{0.1s} = {0.102m/s} }[/math]
Step 5: Update Position (Iteration 1)
Using the position update formula, calculate the position of the mass at the end of the given time step. The average velocity used below is the velocity of the mass at the end of the time step. As with all simplifying assumptions, this measurement is not exact. However, it allows for a close enough approximation that the results derived are still valid.
[math]\displaystyle{ {\vec{r}_{f}} = {\vec{r}_{i} + \vec{v}_{avg}{Δt}} }[/math]
[math]\displaystyle{ {\vec{r}_{f}} = {-1.5m} + {0.102m/s}{0.1s} = {-1.398m} }[/math]
Step 6: Update Time (Iteration 1)
[math]\displaystyle{ {\vec{t}_{f}} = {\vec{t}_{i}} + {Δt} }[/math]
[math]\displaystyle{ {\vec{t}_{f}} = {0s} + {0.1s} = {0.1s} }[/math]
Step 7: Repeat Calculations
Repeat the above calculations using the "Iteration Round 1 Final Values" (calculated above) as the "Iteration Round 2 Initial Values".
Step 8: Update Forces (Iteration 2)
The gravitational force on the mass remains constant, and does not need to be recalculated.
[math]\displaystyle{ {\vec{F}_{spring}} = {-kx} }[/math]
[math]\displaystyle{ {\vec{F}_{spring}} = {-40N/m}\cdot{1.0m-1.398m} = {15.92 N} }[/math]
[math]\displaystyle{ {\vec{F}_{net}} = {\vec{F}_{grav} +{\vec{F}_{spring}}} }[/math]
[math]\displaystyle{ {\vec{F}_{net}} = {-9.8N} +{15.92 N} = {6.12N} }[/math]
Step 9: Update Momentum (Iteration 2)
[math]\displaystyle{ {\vec{p}_{f} = \vec{p}_{i} + \vec{F}_{net}{Δt}} }[/math]
[math]\displaystyle{ {\vec{p}_{f} = {1.02Ns} + {6.12N}\cdot{0.1s} = {1.632Ns}} }[/math]
Step 10: Update Velocity (Iteration 2)
[math]\displaystyle{ {\vec{v}_{f} = \vec{v}_{i} + \frac{\vec{F}_{net}}{m}}{Δt} }[/math]
[math]\displaystyle{ {\vec{v}_{f} = {0m/s} + \frac{1.632N}{1 kg}}{0.1s} = {0.1632m/s} }[/math]
Step 11: Update Position (Iteration 2)
[math]\displaystyle{ {\vec{r}_{f}} = {\vec{r}_{i} + \vec{v}_{avg}{Δt}} }[/math]
[math]\displaystyle{ {\vec{r}_{f}} = {-1.398m} + {0.1632m/s}{0.1s} = {-1.2348m} }[/math]
Step 12: State Answer
After 0.2 seconds, the mass is at a position of 1.23 m below the ceiling, in the vertical direction.
A Note on Iterations: While calculations can be performed manually, as above, it is advisable for more advanced problems that a computer model or similar program be used for such repetitive calculations in order to save time and reduce mathematical errors.
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