Chapter 7. Mass-Spring Systems, Suspension
[First Draft CBPrice October 24th 2011]
The system of a mass coupled with a spring is an interesting model for many real-world situations such as
vehicle suspension, although it is of equal use in understanding how buildings respond to earthquakes or
bridges to pedestrians walking across. These systems show behaviour which we have not yet encountered,
that is oscillations where the position of the mass moves (e.g.) up and down and repeats itself.
Forces on a Mass in a Mass – Spring System.
Take the case of a mass added to the end of a spring, what happens? The mass will move the spring
downwards (since it exerts a downwards force on the spring) but as the spring stretches, it exerts an upwards
force. Eventually these forces balance and so the mass will be at rest. This is shown in Figure 1: In the
leftmost sketch the mass is not connected to the spring, so the spring exerts no force on the mass. In the
centre sketch the mass has pulled the spring down, but the upwards force of the spring on the mass, F is
smaller than the pull of gravity on the mass, so the mass continues to move downwards. In the final figure
the upwards force exerted by the spring on the mass F is equal to the force exerted on the mass downwards
by gravity. The total force is zero so there is no acceleration. This is the equilibrium state.
How does the spring force depend on its “extension”? This has been discovered experimentally, and is
known as “Hooke’s Law” which states that “force is proportional to extension”, (ie double the extension
doubles the force). We can write this mathematically as
Here z is the extension of the spring and k is the “stiffness” of the spring, measured by experiment. A large
value of stiffness means that the spring exerts a larger force for the same extension.
To move to writing simulation code, we know what we must do next. First we write an expression for the
which, using the expressing for the spring force becomes
We can now use the Euler algorithm to calculate the velocity change
and the change in extension is calculated according to
This leads to the following computer code.
velyZ += -1.0*(springK/mmass)*dispZ*dT;
dispZ += velyZ*dT;
Application to a Mountain Bike (rear suspension).
The rear suspension of a mountain bike is shown below in Figure 2 (this is an image extracted from the
associated workshop activity).
There is a pivot near the pedal shaft, around which the front and rear frame components rotate. The spring is
not vertical, but is situated at an angle to the vertical. So when a load is applied to the bike (the rider sits on
it) only a fraction of the load is applied to the spring. But as the frame components rotate around the pivot,
the spring angle changes, which changes the amount of force it exerts. This is a difficult situation to analyse
and simulation can help us understand how the mountain bike suspension behaves.
[Future draft include analysis of the force into components and so help reading of the code].