The formula for the movement of an object attached to a spring is given by the second order linear differential

$mu''(t) + \gamma u'(t) + ku(t) = F(t)$
where
$m$   = mass,
$\gamma$ = damping constant, $k$ = spring constant,
$F ( t )$   is the external force applied to the system, and
$u ( t )$ =   the displacement from equilibrium position at time $t$ of a mass attached to a spring.

In the next few problems, we will focus on determining this differential equation given sufficient information to find $m , \gamma , k$, and $F ( t )$ .  Applications in 3.7 assume no external force (so $F ( t ) = 0)$.

3.7 Weight

The following 2 examples illustrate that in order to determine m, one can be given either mass (m) or weight (mg). Note in our class we will use $m = 9.8m/sec^2\hbox{ or } 32 ft/sec^2$.

Example 1: A spring mass system has a spring constant = 3N/m. A mass of 4kg is attached to the spring. If there is no damping and no external force, state the 2nd order differential equation that models the location of the mass.

Example 2: A spring mass system has a spring constant = 3 lbs. A mass weighing 4 lbs is attached to the spring. If there is no damping and no external force, state the 2nd order differential equation that models the location of the mass.

NOTE: normally you will also be given initial values. See later examples.

3.7 Damping

Damping, like friction, will slow the movement of the object over time. Thus as we will see later, when there is damping, $u(t)$ goes to 0 as $t$ goes to $\infty$. Recall that the damping force $F_{damping}= -\gamma u'(t)$. Thus if you know the damping force and velocity at a particular time, you can determine $\gamma$. Velocity, $u'(t)$, can be positive or negative, but since you know $\gamma\geq 0$, you do not need to pay attention to signs (+/-) as long as you take $\gamma$ to be positive when there is a damping force.

Example 1: A mass of 4kg is attached to the spring with spring constant $k$ = 3N/m. The mass is also attached to a viscous damper with a damping constant of 6N sec/m. If there is no external force, state the 2nd order differential equation that models the location of the mass.

Example 2: A mass of 4kg is attached to the spring with spring constant k = 3N/m. The mass is also attached to a viscous damper that exerts a force of 10N when the velocity of the mass is 2m/sec. If there is no external force, state the 2nd order differential equation that models the location of the mass. Another way to word example 2 is Example reworded 2: A mass of 4kg is attached to the spring with spring constant k = 3N/m. The mass moves in a medium that imparts of viscous force of 10N when the speed of the mass is 2m/sec. If there is no external force, state the 2nd order differential equation that models the location of the mass.

3.7 Spring constant

The following 2 examples illustrate how to determine $k$ = spring constant. Recall that the force from the spring is balance by the force from gravity. Thus $kL = mg$ where $L$ is the length that the spring is stretched by attaching an object with mass $m$ (or weight $mg$).

Example 1: A 20 kg mass stretches a spring 2 m. If there is no damping and no external force, state the 2nd order differential equation that models the location of the mass.

Example 2: A spring is stretched 2m by a force of 100N. A mass of 4kg is attached to the spring. If there is no damping and no external force, state the 2nd order differential equation that models the location of the mass.

3.8 External force

In section 3.8, an external force will be applied to the system. Be careful to determine whether the force(s) given in the problem refer to an external force or a damping/viscous force or spring force.

Example: A spring is stretched 2m by a force of 100N. A mass of 4kg mass is attached to the spring and acted on by an external force of $21sin(9t)$ N. The mass is also attached to a viscous damper that exerts a force of 10N when the velocity of the mass is 2m/sec. State the 2nd order differential equation that models the motion of the mass.

Initial values

We will now focus on setting up the initial value problem (not just the differential equation per previous examples).

Note: To state the initial value problem, you must state both the differential equation and the initial value(s).

In the case of mechanical (and electrical) vibrations, the differential equation is 2nd order. Thus we must specify a point that the solution passes thru as well as the slope of the tangent line at that point.

Thus for mechanical vibrations, the IVP is $mu''(t) + \gamma u'(t) + ku(t) = F(t), ~~ u(t_0) = u_0, ~ u'(t_0) = v_0$.

Thus at time $t_0$, the mass is at location $u_0$ with velocity $v_0$.

In the graph of $u$, the mass passes thru the point $(t_0, u_0)$ and the slope of the tangent line at this point is $v_0$.

Sidenote: Compare to calculus 1 word problems where you needed to do 2 integrals to solve the problem. Two integrals means 2 constants. To determine these 2 constants in calculus 1, you were given the point that the solution passed thru and the slope of the tangent line at this point (for example, the height from which a ball was thrown and the ball's initial velocity).

Initial conditions part 1

Note: the positive direction points down.

Thus if the spring is pulled down, $u > 0$. If it is compressed, $u < 0$. If the mass starts at the equilibrium position $u(0) = 0$

If the mass is moving down in the positive direction, then $u' > 0$.
If the mass is moving up in the negative direction, then $u' < 0$.
If the mass is simply released and not given an initial velocity, $u'(0) = 0$.

In the following examples, $m = 1$, $k = 4$, and there is no damping. State the initial value problems that model the following scenarios.

Suppose $m = 1$, $k = 4$, and there is no damping. The mass is pulled down 2 meters from its equilibrium position and set in motion with a downward velocity of 3 m/sec.

Suppose $m = 1$, $k = 4$, and there is no damping. The mass is pulled down 2 meters from its equilibrium position and set in motion with an upward velocity of 3 m/sec.

Suppose $m = 1$, $k = 4$, and there is no damping. The mass is pulled down 2 meters from its equilibrium position and then released.

Suppose $m = 1$, $k = 4$, and there is no damping. The mass is set in motion from its equilibrium position with an upward velocity of 3 m/sec.

Initial conditions part 2

To repeat for emphasis: Note, the positive direction points down.

Thus if the spring is pulled down, $u > 0$. If it is compressed, $u < 0$. If the mass starts at the equilibrium position $u(0) = 0$

If the mass is moving down in the positive direction, then $u' > 0$.
If the mass is moving up in the negative direction, then $u' < 0$.
If the mass is simply released and not given an initial velocity, $u'(0) = 0$.

In the following examples, $m = 1$, $k = 4$, and there is no damping. State the initial value problems that model the following scenarios.

Suppose $m = 1$, $k = 4$, and there is no damping. The mass is pushed up 2 meters from its equilibrium position and set in motion with an upward velocity of 3 m/sec.

Suppose $m = 1$, $k = 4$, and there is no damping. The spring is compressed 2 meters and then released.

Suppose $m = 1$, $k = 4$, and there is no damping. The mass is pushed up, contracting the spring 2 meters and set in motion with a downward velocity of 3 m/sec.

Initial conditions part 3 State the initial value problem that models the following scenario:

A 3 kg object is attached to a spring and will stretch the spring 392 mm by itself. There is no damping in the system and an external force of the form F(t)=10cos(5t) is applied to the object. If the object is initially displaced 20 cm downward from its equilibrium position and given a velocity of 10 cm/sec upward find the displacement at any time t. Note this problemis based on example 5 from Paul's online notes