Free Undamped Motion of Spring/Mass system
Recall from Calculus I, Hooke's Law for springs, F
= kx, where F is the force, k is the spring constant and x is the
displacement beyond equilibrium. Then we found the work done in
stretching or compressing the spring by integrating the force over
this distance. What we are to do now is find the equation of motion
of the spring after it is stretched or compressed and then released.
For free undamped motion where there is no resistance from the
surroundings, the differential equation 
,
where k is the spring constant, and m the mass( use slugs, not
pounds, in the Engish system ) of the object at the end of the
spring, is solved for x[ t ] the displacement of the spring
at any time t. In these problems, displacement below
equilibrium is positive. Also, the constant
,
is used.
For example, if a spring has k = 8, m = 2, x( 0 ) = 4, x'( 0 ) = 6, the differential equation to be solved is x''( t ) + 4x( t ) = 0, x( 0 ) = 4, x'( 0 ) = 6. Here the mass m is 2 slugs which means the weight is 64 pounds. The x( 0 ) = 4 is the position of the weight at t = 0 which is 4 units below equilibrium length. The initial velocity x'( 0 ) = 6 means that the spring was given a push downward at 6 ft/sec. The solution obtained from DSolve[{ x''[ t ] + 4x[ t ] == 0, x[ 0 ] == 4, x'[ 0 ] == 6 }, x[ t ], t] is x[ t ] = 4 Cos[ 2t ] + 3 Sin[ 2t ]. Its graph is:
The above graph plots the motion of the displacement of the spring with respect to equilibrium over time. The actual motion of a spring of length 10 ft, with an equation of motion x( t ) = 10 + 2sin( 2t ) - 6cos( 2t ) is:
Free Damped Motion
When the surroundings offer resistance to the
spring, damped motion results. It is clear from the above solution
that undamped is unrealistic as the motion of the spring will not
last forever. For damped motion, a damping constant,
,
is the constant proportional to the velocity of the spring. The
resulting second order linear equation is:
,
where 
.
There are three possibilities here:
In the overdamped or critically damped cases, the spring moves through equilibrium at most one time. In an underdamped situation the spring moves through equilibrium two or more times.
For example, if m = 1, k = 10,
= 2, x'( 0 ) = 2, x( 0 ) = 4, the differential equation is: x''( t )
+ 2x'( t ) + 10x( t ) = 0 and is underdamped. The solution is x( t )
= e( -t )( 4cos( 3t ) + 2sin( 3t ) ) with the
graph:
Driven Motion
If the spring is attached to an external force, f(
t ), the differential equation becomes nonhomogeneous. Now we must
solve:
.
Let's apply a force of f( t ) = 6sin( t ) to the above system and
solve. The solution is
.
It is agood thing that Mathematica is doing all the work! Please note
that the first part will go to 0 as t increases. This part is the
transient part while the second part is the steady state. Here is the
graph:
LRC Circuits
The spring/mass systems are analagous to LRC
circuits, where the transient and steady state terms are derived. The
differential equation to be solved is 
,
where the constants are precisely what we have already
done.
Things to know and do
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