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]
[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 because dissipative forces prevalent such as air resistance, friction, fluid drag, etc. reduce the net energy of the system by dissipating it into the surroundings. A formal derivation of the equations of motion for a damped system will not be considered here, but it is very straightforward to add damping to our computational model. Whenever we calculate the net force on the mass during the iterative process, all we have to do now is add a dissipative force. For air resistance, the standard formula is
[math]\displaystyle{ \vec{F} = -0.5CAρv^2\hat{v} }[/math]
More frequently though, when we talk about the damped harmonic oscillator, we will discuss this force[1]:
[math]\displaystyle{ \vec{F} = -b\vec{v} }[/math]
where b is the damping constant. The important difference between the two is that in the latter the force has a linear relationship to [math]\displaystyle{ \vec{v} }[/math]. There are three types of damping, depending on the coefficient. If the coefficient is small relative to the mass and spring constant, then the oscillator will be underdamped, and still have some oscillatory behavior. If it is large relative to the mass and spring constant, the oscillator will be overdamped, meaning that there will be no oscillations, but the system will take longer to arrive at an equilibrium because it has been slowed down by the dissipation. The fastest time of decay is when the system is critically damped, which occurs when [math]\displaystyle{ b^2 = 4mk }[/math]. The figure below illustrates these three possibilities, and is generated using the numpy based code from the Computational Methods section below.
Gravity
Introducing gravity to our model of spring mass systems is fairly straightforward (actually solving for it is somewhat more complicated, and we will only do so with computational methods described below). One simply adds the standard gravitational force [math]\displaystyle{ F = mg }[/math] to the force equations one has already. The most important point is to always remember which way is down: if your spring is hung upside down, then you may have chosen to define the origin at the equilibrium point and [math]\displaystyle{ +x }[/math] as being the downward direction, in which case you must take care to ensure that the sign of gravity corresponds.
A Computational Model
By definition, iterative prediction requires repeated calculations. This is extremely tedious to do by hand, so it is greatly preferable to perform them with a computer. Two programs are presented which perform these calculations.
An implementation of iterative motion for a spring system which feels gravitational force is provided in VPython here. The operative portion of the code is copied in the collapsible below (note, if one wishes to use this code, it will be necessary to include the proper import statements, which may be found at the link).
## constants and data
g = 9.8
mball = .5 ## mass of ball
L0 = 0.3 ## relaxed length of spring
ks = 50 ## spring constant (in N/m)
deltat = .001 ## time step
fscale = .07 ## scale factor for force arrow
pscale = .4 ## scale factor for position arrow
t = 0 ## initial time = 0 seconds
##########
## objects
ceiling = box(pos=vec(0,0,0), size=vec(0.2,0.01,0.2)) ## origin is at ceiling
ball = sphere(pos=vec(-.2637,-.1047,-.0429), radius=0.025, color=color.orange) ## note: spring initially compressed
spring = helix(pos=ceiling.pos, color=color.cyan, thickness=.003, coils=40, radius=0.015) ## create spring
spring.axis = ball.pos - ceiling.pos
trail = curve(color=ball.color) ## create trail of ball's position
fnetarrow = arrow(pos = ball.pos, color = color.blue) ## create arrow showing vector net force on ball
parrow = arrow(pos = ball.pos, color = color.green) ##create arrow showing vecotr momentum of ball
##########
## initial values
ball.p = mball*vector ## calculate initial momentum of ball
#########
## improve the display
scene.autoscale = 0 ## don't let camera zoom in and out as ball moves
scene.center = vector(0,-L0,0) ## move camera down to improve display visibility
##########
## calculation loop
while t < 9.02: ## sets end time for loop
rate(300)
## calculate force on ball by spring
l = ball.pos - ceiling.pos
s = l.mag - L0
lhat = l/l.mag
fspring = ks * s * -lhat
## calculate net force on ball
fnet = fspring + mball*vector(0, -g, 0)
fnetarrow.pos = ball.pos
fnetarrow.axis = fnet * fscale
## apply momentum principle
ball.p = ball.p + fnet*deltat
parrow.pos = ball.pos
parrow.axis = ball.p * pscale
## update position
ball.pos = ball.pos + (ball.p/mball)*deltat
## update axis of spring
spring.axis = ball.pos - ceiling.pos
trail.append(pos=ball.pos)
## update time
t = t + deltat
############
## Produce results
print(ball.pos)
print(ball.p/mball)
A second implementation using Numpy may be found here.
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 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: Preliminaries It is first necessary to determine preliminaries. Since this is a one dimensional problem, no vectors are needed, but it is important to properly define our signs. Here we will assume the up is positive, so [math]\displaystyle{ F_g = mg }[/math] with [math]\displaystyle{ g = -9.8 m/s^2 }[/math]. Additionally, we must choose a time step: 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. It is appropriate in this situation to take [math]\displaystyle{ 0.1 s }[/math] as our time step.
Step 2: Calculating 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 spring force. The net force may thus 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}} = m\vec{g} }[/math]
[math]\displaystyle{ {\vec{F}_{grav}} = {10kg}\cdot(-9.8 m/s^2) = {-9.8 N} }[/math]
[math]\displaystyle{ {\vec{F}_{spring}} = {-kx} }[/math]
[math]\displaystyle{ {\vec{F}_{spring}} = {-40N/m}\cdot((-1.5 m)-(-1.0 m)) = {20 N} }[/math]
Note how the gravitational force is negative (downwards, as expected) and the spring force is positive (upwards, in opposition to gravity, as expected).
[math]\displaystyle{ {\vec{F}_{net}} = {\vec{F}_{grav} +{\vec{F}_{spring}}} }[/math]
[math]\displaystyle{ {\vec{F}_{net}} = {-9.8N} +{20 N} = {10.2 N} }[/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\Delta t }[/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}{Δt}} }[/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
Predicting Change in multiple dimensions
Fundamentals of Iterative Prediction with Varying Force
Length and Stiffness of an Interatomic Bond
External links
Interactive Spring-Mass Simulator