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 page Fundamentals of Iterative Prediction with Varying Force lays out the general methods of iterative prediction, but a brief recapitulation is provided here. he 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{ \Delta p = \vec{F}_{net}\Delta t }[/math]

[math]\displaystyle{ \vec{p}_{f} - \vec{p}_{i} = \vec{F}_{net}\Delta t }[/math]

[math]\displaystyle{ \vec{p}_{f} = \vec{p}_{i} + \vec{F}_{net}\Delta t }[/math]

Kinematics, expressed in a similarly iterative manner, allows for the determination of the next iteration's position and velocity

[math]\displaystyle{ \vec{v}_{f} = \frac{\vec{p}_f}{m} }[/math]

[math]\displaystyle{ \vec{r}_{f} = \vec{r}_{i} + \vec{v}_{avg}\Delta t }[/math]

Damping

We know from our observations of say a pendulum swinging or a spring bobbing that it's periodic motion eventually dies out. This is due to the dissipative forces prevalent in nature (air resistance and viscous drag ) that dissipate the potential and kinetic energies of the system (spring+mass) to the surroundings (everything outside our chosen system ) into other forms of energy (heat, sound, etc.). When this happens, we say that the system is executing damped oscillations.

It takes some hairy math to arrive at an approximate analytical solution for the above case. Instead of doing that, however, it is much easier for us to encode the presence of dissipative forces if we were to carry out numerical integration using an iterative process.

Whenever we calculate the net force on the mass during the iterative process, all we have to do now is add a force that is representative of any of the dissipative forces. If we are dealing with air resistance, then we can use the equation

[math]\displaystyle{ \vec{F} = -0.5CAρv^2\hat{v} }[/math]

Else, we can use this equation

[math]\displaystyle{ \vec{F} = -b\vec{v} }[/math] where b is the damping constant

A Computational Model

By definition, iterative prediction requires repeated calculations. Computations that require significant time and attention to detail when performed by hand can be done incredibly quickly and accurately by a computer model.

Coding iterative prediction is very similar to writing equations by hand. The same approach is used; the largest difference arises in syntax, as coding requires variables and calculations to be written in a specific format that varies depending on the program used. In the following discussion, VPython will be used to illustrate a computational model of iterative prediction of a spring-mass system.

The text in italics is valid Python code. Notes in regular text to the right the hashtag (#) on each line are explanatory notes that are not part of the code. The complete program, which includes some visual aspects not described below, can be viewed here.

1) Determine initial parameters and constants

The required variables will vary depending on the problem but a general example is provided below:

 g = 9.8            ##gravitational constant

 mball = .5         ##mass of ball

 L0 = 0.3           ##relaxed length of spring

 ks = 50             ## spring constant (in N/m)

 deltat = .001     ## time step

 t = 0               ## initial time = 0 seconds

 velocity = (-0.1, 0.2, 0.4)

2) Create objects

Create visual representations of the spring and mass attached to it

 ceiling = box(pos=vec(0,0,0), size=vec(0.2,0.01,0.2))

The initial position of the ball is indicated here. An arbitrary size and color have been assigned.

 ball = sphere(pos=vec(-.1,-.1,-.1), radius=0.025, color=color.orange)

The spring is attached to the ceiling, as indicated by "pos=ceiling.pos". Again, an arbitrary color and size, as well as number of coils have been assigned to it.

 spring = helix(pos=ceiling.pos, color=color.cyan, thickness=.003, coils=40, radius=0.015)

The spring should extend from the point of attachment (ceiling,pos) to the position of the ball (ball.pos), so the next line of code should be included as well.

 spring.axis = ball.pos - ceiling.pos

3) Calculate initial momentum

Find the initial momentum by calculating the product of the velocity vector and mass of the ball.

ball.p = mball*velocity

4) Begin calculation loop

This will include all equations that must be repeated. Begin by specifying a terminating condition for the loop. This usually consists of an end time, but does not have to.

while t < 10

5) Calculate initial forces acting on system, and sum to find net force

Calculate the spring force using Hooke's Law:

l = ball.pos - ceiling.pos       ## calculate the length of the spring

s = l.mag - L0                        ## stretch (s) is equal to the difference in the length of the spring (l) and the relaxed length (L0)

lhat = l/l.mag                         ## determine L(hat)

fspring = ks * s * -lhat          ##apply Hooke's Law

Calculate the gravitational force

fgrav = mball*vector(0, -g, 0)

Calculate the net force

fnet = fspring + fgrav

6) Update object's momentum

Apply the momentum principle

ball.p = ball.p + fnet*deltat

5) Update object's velocity

ball.v = ball.p/mball

6) Update object's position

ball.pos = ball.pos + ball.v*deltat

7) Update time

t = t + delta t

8) Repeat calculations

VPython will continue iterating through the calculations included in the loop until the terminating condition is met.

For complete visual code, click here.

Examples

Simple Example

vertical spring-mass system

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 10 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.

A Note on Iterations: While calculations can be performed manually, as above, it is advisable for more advanced problems that VPython or a similar program be used for such repetitive calculations in order to save time and reduce the likelihood of mathematical errors.

Middling Example

A semi-challenging problem, such as the one below, will often require you to perform vector calculations. However, the same steps as the easy example above still apply. Pay careful attention to signs as you solve these types of problems.

The simplest example of a spring mass system is one that moves in only one-direction.

Consider a massless spring of relaxed length 0.5 m with spring constant 120 N/m attached to the ceiling. If a 1.4 kg mass is released with an initial velocity of <1.2, 0.8, -0.2> m/s at an initial position of <-0.3, 0.7, 0.2> m, what is the velocity of the block after 0.005 seconds? Consider the point of attachment on the spring to the ceiling to be the origin. Only one iteration is necessary.

Step 1: Calculate Initial Momentum The initial momentum is simply the product of the mass and velocity of the object.

[math]\displaystyle{ {\vec{p}_{i}} = {{m}\cdot\vec{v}_{i}} }[/math]

[math]\displaystyle{ {\vec{p}_{i}} = {{1.4kg}\cdot{\lt 1.2, 0.8, -0.2\gt m/s}} = {\lt 1.68, 1.12, -0.28\gt Ns} }[/math]

Step 2: Calculate Force of Gravity

[math]\displaystyle{ {\vec{F}_{grav}} = {m}{\lt 0, -g, 0\gt } }[/math]

[math]\displaystyle{ {\vec{F}_{grav}} = {1.4kg}\cdot{\lt 0, -9.8, 0\gt m/s/s} = {\lt 0, -13.72, 0\gt N} }[/math]

Step 3: Calculate Force of Spring

[math]\displaystyle{ {\vec{F}_{spring}} = {-k}{(\vec{L}_{mag}-\vec{L}_{0})} \cdot {\hat{L}} }[/math]

The magnitude of L is equal to the square root of the squares of it's individual components

[math]\displaystyle{ {\hat{L}} = {\sqrt{ {{L}_{x}}^{2} + {{L}_{y}}^{2} + {{L}_{z}}^{2} }} }[/math]

[math]\displaystyle{ {\hat{L}} = {\sqrt{ {-0.3}^{2} + {0.7}^{2} + {0.2}^{2} }} = {\sqrt{0.62}} }[/math]

The unit vector of L ([math]\displaystyle{ {\hat{L}} }[/math]) is equal to L divided by its magnitude

[math]\displaystyle{ {\lVert\vec{L}\rVert} = {\frac{\vec{L}}{{\lVert\vec{L}\rVert}}} }[/math]

[math]\displaystyle{ {\lVert\vec{L}\rVert} = {\frac{\lt -0.3, 0.7, 0.2\gt m }{\sqrt{0.62}}} }[/math]

Substituting these values into Hooke's Law, then gives the force exerted by the spring:

[math]\displaystyle{ {\vec{F}_{spring}} = {-k}{(\lVert\vec{L}\rVert-\vec{L}_{0})} \cdot {\hat{L}} }[/math]

[math]\displaystyle{ {\vec{F}_{spring}} = {-120 N/m}{(\sqrt{0.62} m - {0.5 m})} \cdot {\frac{\lt -0.3, 0.7, 0.2\gt m }{\sqrt{0.62}}} }[/math]

[math]\displaystyle{ {\vec{F}_{spring}} = {\lt 12.83316, -29.94405, -8.55544\gt N} }[/math]

Step 4: Caculate Net Force"

The net force can be calculated by summing these two forces.

[math]\displaystyle{ {\vec{F}_{net}} = {\vec{F}_{grav} +{\vec{F}_{spring}}} }[/math]

[math]\displaystyle{ {\vec{F}_{net}} = {\lt 0, -13.72, 0\gt N} +{\lt 12.83316, -29.94405, -8.55544\gt N} = {\lt 12.83316, -16.22405, -8.55544\gt N} }[/math]

Step 5: Update Momentum

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}{&Delta;t}} }[/math]

[math]\displaystyle{ {\vec{p}_{f} = {\lt 1.68, 1.12, -0.28\gt Ns} + {\lt 12.83316, -16.22405, -8.55544\gt N}\cdot{0.005s} = { Ns}} }[/math]

[math]\displaystyle{ {\vec{p}_{f} = {\lt 1.7442, 1.0389 ,-0.3611\gt Ns}} }[/math]

Step 4: Calculate Velocity

[math]\displaystyle{ {\vec{v}_{f}} = \frac{\vec{p}_{f}}{m} }[/math]

[math]\displaystyle{ {\vec{v}_{f}} =\frac{\lt 1.7442, 1.0389 ,-0.3611\gt Ns}{1.4kg} = {\lt 1.2458, 0.7421, -0.2579\gt m/s} }[/math]

Step 5: State Answer

After 0.005 seconds, the mass is moving with velocity <1.25, 0.74, -0.26> m/s

Difficult Example

Very difficult iterative motion problems likely require several calculation loops to be performed. Thus, it is recommended that a computer program, such as VPython, is used to solve these. For general instructions on how to do this, please see above in this article, under the heading A Computational Model.

Connectedness

While you may not need to precisely calculate the motion of spring-mass systems on a daily basis, understanding their behavior has led to the invention of many useful items. Among these are mattresses, clocks, and the suspension on cars.

However, the concept of iterative prediction applies to far more than masses on springs, as the concepts of physics apply far beyond textbook problems. By thinking in small enough time increments, the behavior of any system can be understood.

For example, in chemical engineering, entire processes can be broken down into simpler pieces and analyzed. These processes could then be modeled in order to predict future behavior.

See also

Conservation of Momentum

Predicting Change in Multiple Dimensions

Derivation of Momentum Principle

Momentum Principle

Analytical Prediction

Net Force

Young's Modulus

Tension

Fundamentals of Iterative Prediction with Varying Force

Simple Harmonic Motion

Length and Stiffness of Interatomic Bond

External links

Interactive Spring-Mass Simulator

References