This simulation shows a single mass on a spring, which is connected to a wall. This is an example of a simple linear oscillator.

Enable the "show controls" checkbox to change mass, spring stiffness, and friction (damping). You can drag the mass with your mouse to change the starting position.

(If you don't see the simulation try instructions for enabling Java.)

- What is the relationship between acceleration and position?
- How do mass or spring stiffness affect the relationship between acceleration and position?
- How do mass or spring stiffness affect the period or frequency of the oscillation?

To show the graph and parameters, click on the "show graph+controls" button above. You can set what variables are shown on the vertical or horizontal axis by selecting from the pop-up menu above. To change a parameter, click on it, type the new value and hit the enter key.

*x*= position of the block*v*=*x*' = velocity of the block*m*= mass of the block*R*= rest length of the spring*k*= spring stiffness*b*= damping constant (friction)

*F*_{spring} = −*k* × stretch

*F*_{spring} = − *k* *x*

*F* = *F*_{spring} + *F*_{damping} = − *k* *x* − *b* *v*

*m* *x*'' = −*k* *x* − *b* *v*

x'' = − ^{k}⁄_{m} x − ^{b}⁄_{m} x'
| (1) |

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 = −

*x*' = *v*

*v*' = − ^{k}⁄_{m} *x* − ^{b}⁄_{m} *v*

To begin the simulation, we initialize the two variables

With no damping (

where

The frequency is the inverse of the period:

frequency =

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

*x*'' = − ^{k}⁄_{m} *x*

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

Answer: From the equation

*x*'' = − ^{k}⁄_{m} *x*

- 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?

Answer: The analytic solution is

and the frequency is given by

frequency =

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;

TO DO:

- Back to simple graph (without overlapping timelines). Or? Maybe keep that graph to demonstrate the meaning of phase graph?
- Fix the energy graph to show work done
- References: Give links to related Wikipedia or other pages
- Rewrite the intro...

This web page was first published April 2001.