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

$m$ = 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