Motion of Springs
Theorem 1 (Hook’s Law). A spring stretched or compressed x units from its natural length exerts a force (called the
restoring force) proportional to x:
restoring force = −kx.
(Why the negative? From Newton’s Third Law they must be equal [in magnitude] and opposite [in direction].)
Undamped Motion
If we let
x(t) = the position of the mass at time t
then
acceleration =
and
F = ma = m
d2x
dt2
d2x
.
dt2
(Note: You need to convert lbs to a unit of mass (slug) by dividing by 32 sft2 when given English units.)
This means that
m
d2x
= −kx
dt2
or
d2x
+ kx = 0
dt2
is the differential equation that models the undamped motion of a mass on a spring. The solution family will always have
the form
x(t) = c1 cos(ωt) + c2 sin(ωt).
m
Damped Motion (Undriven)
Now suppose we add a damping force to the system (such as putting the spring in a fluid, attaching a shock absorber,
or accounting for friction in the system). In this model we will assume that this damping force is proportional to the
velocity of the object. This yields the DE
d2x
dx
m 2 = −kx − β
dt
dt
or
d2x
dx
m 2 +β
+ kx = 0.
dt
dt
There are three possible forms of the solution family, corresponding to the three different possibilities for the roots of
the auxiliary equation of this second order, constant coefficient, linear equation:
1
Overdamping
If there are two distinct real roots m1 , m2 then x(t) = c1 em1 t + c2 em2 t and the graph will have the shape shown in
Figure 1. This is called “overdamping.”
Figure 1: Overdamped motion
Critical damping
If there is one repeated real root m then x(t) = c1 em t + c2 tem t and the graph will have the shape shown in Figure 2.
This is called “critical damping.”
Figure 2: Critically damped motion
Underdamping
If there are two complex conjugate roots α ± βi then x(t) = eαt (c1 cos(βt) + c2 sin(βt)) and the graph will have the shape
shown in Figure 3. This is called “underdamping.”
Figure 3: Underdamped motion
Driven Damped Motion
We now add a driving force f (t):
m
d2x
dx
= −kx − β
+ f (t)
dt2
dt
or
d2x
dx
+β
+ kx = f (t).
dt2
dt
In this case we would find x(t) = xc (t) + xp (t). If parts of xc are transient (which will happen when there is a negative
exponent in the exponential terms) then that part will approach 0 as t → ∞ and leave us with a steady state solution.
m
2
Examples
1. A spring with a 4 kg mass and natural length 1 m is maintained stretched to a length of 1.3 m by a force of 24.3
N. If the spring is compressed to a length of 0.8 m and released with 0 velocity then find the position of the mass
at time t.
2. A mass weighing 3 pounds stretches a spring 18 inches. At t = 0 the mass is released from equilibrium with an
upward velocity of 32 ft/s. Determine the equation of motion.
3. A spring with a mass of 2 kg has damping constant 8 and spring constant 80. Determine whether the mass
is overdamped, critically damped or underdamped. Find the position of the mass at time t if it starts at the
equilibrium position with a velocity of 4 m/s.
4. A mass of 2 kg is attached to a 0.5 m spring. A force of 25.6 N is required to maintain it stretched to a length
of 0.7 m. Suppose the spring is immersed in fluid with damping constant c = 40. Determine whether the mass is
overdamped, critically damped or underdamped. Find the equation of motion if the mass starts from the equilibrium
position and is given a push downward to start it with an initial velocity of 0.6 m/s.
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