# Einzelne Feder

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Diese Simulation zeigt eine einzelne Masse an einer Feder, die Feder ist mit der Wand verbunden. Dies ist ein Beispiel von einer einfachen linearen Oszillator.

Sie können wechseln Masse, Feder Steifigkeit, und Dämpfung (Reibung). Sie können die Masse schleppen mit eure Maus um die Anfangsposition zu ändern.

The math behind the simulation is shown below. Also available are: open source code, documentation and a simple-compiled version which is more customizable.

## Rätsel

Versuchen Sie den Graphen zu benutzen um mit hilfe wechselnder Parameter wie Masse oder Federsteifigkeit die Antworten zu den folgenden Fragen zu erhalten.

• Was ist das Verhältnis zwischen Beschleunigung und Position?
• Wie kann Masse oder Federsteifigkeit das Verhältnis zwischen Beschleunigung und Position verändern?
• Wie kann Masse oder Federsteifigkeit die Periode oder Frequenz der Oszillation verändern?

Sie finden die Antworten unten.

## Physik

Wir definieren die folgene Variablen und Konstanten

• x = Position des Blockes
• v = x' = Geschwindigkeit des Blockes
• m = Masse des Blockes
• R = Federlänge im Ruhestand
• k = Federsteifigkeit
• b = Dämpfung (Reibung) Konstante

Eine Feder entwickelt eine Kraft proportional dazu, wie weit sie ausgestreckt ist. Die Kraft wirkt in gegenteilige Richtung der Ausdehenung.

Fspring = −k × stretch

Wenn wir unser Koordinatesystem einrichten so dass x = 0 entspricht, sodass die Feder nicht ausgestreckt ist, dann ist die Streckung der Feder gleich x . Die Gleichung für die Kraft der Feder ist:

Fspring = − k x

In addition, there is a damping (friction) force that resists the motion. It is proportional to the velocity. So we add Fdamping = −b v to get the total force

F = Fspring + Fdamping = − k xb v

Combining this with Newton's law of motion F = m a , and the definition of acceleration as the second derivative of position a = x'' we have the differential equation:

m x'' = −k xb v

or equivalently:

 x'' = − k⁄m x − b⁄m x' (1)

Equation (1) is the equation of motion for the spring, defining exactly what happens over time.

You can actually see equation (1) visually if you turn on the graph in the simulation. Suppose there is no damping (set damping to zero). Then if you set the graph on the simulation to plot acceleration against position, you will get a straight line, with slope = −k/m . So if you increase the stiffness of the spring, the line becomes steeper. If you increase the mass, the line becomes less steep.

## Numerical Solution

To solve this equation numerically (ie. by computer) we use the Runge-Kutta method. To do so we need to convert the second order differential equation (1) into a set of first order differential equations. Note that we can write the acceleration as the first derivative of velocity: x'' = v' . Therefore we can express equation (1) as a system of two first order differential equations:

x' = v

v' = − km xbm v

This is the form that we need in order to use the Runge-Kutta method for numerically solving the differential equation.

To begin the simulation, we initialize the two variables x,v for their value at time t=0 . Then we use the Runge-Kutta algorithm to calculate the values of x,v after a short time interval, and this continues indefinitely.

## Analytic Solution

An analytic solution uses mathematics to find the solution instead of the brute force of the computer. The advantage is that we can get some more insight from the analytic solution, instead of having to analyze zillions of numbers that come from a numerical solution.

With no damping ( b = 0 ) and the block initially not moving, the analytic solution is given by

$$x(t) = x_0 \cos(\sqrt{k/m} \; t)$$

where x0 = initial position of the block and t = time. The period of the oscillation is the time it takes to repeat. From the solution we see that the oscillation repeats when $$\sqrt{k/m} \; t = 2 \pi$$, and so the period is

$$t = 2 \pi \sqrt{m/k}$$

The frequency is the inverse of the period:

$$frequency = \frac{1}{2 \pi} \sqrt{k/m}$$

So we predict that

• increasing mass by 4 times doubles the period and halves the frequency;
• increasing spring stiffness by 4 times halves the period and doubles the frequency;

You can check these predictions by modifying the parameters on the simulation (you'll need a stopwatch to time the frequency).

You can view a page with the derivation of the analytic solution.

What is the relationship between acceleration and position?

Answer: It is a linear relationship as given by the equation

x'' = − km x

where x = position, x'' = acceleration, m = mass, and k = spring stiffness.

How do mass or spring stiffness affect the relationship between acceleration and position?

x'' = − km x

we know that there is a simple linear relationship between acceleration and position. The mass and spring stiffness affect the slope of this line.

• Increasing mass makes the line less steep.
• Increasing spring stiffness makes the line steeper.

How do mass or spring stiffness affect the period or frequency of the oscillation?

$$x(t) = x_0 \cos(\sqrt{k/m} \; t)$$

and the frequency is given by

$$frequency = \frac{1}{2 \pi} \sqrt{k/m}$$

So we predict that

• increasing mass by 4 times doubles the period and halves the frequency;
• increasing spring stiffness by 4 times halves the period and doubles the frequency;

This web page was first published April 2001.

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