11.7 Applications for Mass Spring Chain

Types: N/A
Examples: N/A
Constructions: N/A
Generalizations: N/A

Properties: N/A
Sufficiencies: N/A
Questions: N/A

Recall 3.1.2 Vector Model of Mass-Spring Chain.

Elongation Vector

Let's derive a vector to represent the "stretch" of the springs.

\begin{proof}[Solution.]
To start, let's look at spring 1.

$e_{1}(t)$ is the amount of "net" stretch of spring 1. Therefore, $$ e_{1}(t) = u_{1}(t), $$ which is the displacement value of mass 1. - & Recall that displacement is the total change in position from the origin.

Now, let's look at spring 2.

Suppose $m_{1}$ had 0 displacement. $$ \implies u_{1}(t) = 0 $$ How much would $m_{2}$ drop? $$ \begin{align*} e_{2}(t) &= \text{amount of "net" stretch of spring 2 at time }t\\ &= u_{2}(t)-u_{1}(t) \end{align*} $$

Remark. Hooke's law is more about elongation than displacement. This is because we are assuming the frame of reference is at a fixed point ( $m_{1}$ does not move). In other words, the possible movement of the top spring is not factored in.
$⧫$

Lastly, let's look at the bottom spring, spring 3.

$$ \begin{align*} e_{3}(t) &= \text{elongation of spring 3 at time }t\\ &= \underbrace{ -u_{2}(t) }_{ \text{negative of displacement} } \end{align*} $$ - When elongation is negative, the spring length is shorter than it's original length.

Remark. When $u_{2}$ is positive, the spring is stretched and vice versa.
$⧫$

Now that we have our elongation values, we can create our elongation vector

\vec{e} (t) = [\begin{matrix} e_{1} (t) \\ e_{2} (t) \\ e_{3} (t) \end{matrix}] = [\begin{matrix} u_{1} (t) \\ u_{2} (t) - u_{1} (t) \\ - u_{2} (t) \end{matrix}]

Can we rewrite this as a 6.1 Linear Combination of Vectors?

\begin{aligned} = [\begin{array}{c} u_{1} (t) + 0 \\ u_{2} (t) - u_{1} (t) \\ 0 - u_{2} (t) \end{array}] \\ = [\begin{array}{c} u_{1} (t) \\ - u_{1} (t) \\ 0 \end{array}] + [\begin{array}{c} 0 \\ u_{2} (t) \\ - u_{2} (t) \end{array}] \\ ⟹ \vec{e} (t) & = \underset{scalar \times column vector}{\underset{⏟}{u_{1} (t) [\begin{array}{c} 1 \\ - 1 \\ 0 \end{array}]}} + \underset{scalar \times column vector}{\underset{⏟}{u_{2} (t) [\begin{array}{c} 0 \\ 1 \\ - 1 \end{array}]}} \end{aligned}

Recall the definition of 10.1 Matrix-Column-Vector Multiplication via Linear Combinations. We can rewrite our elongation vector as a matrix vector multiplication problem.

\begin{aligned} = {[\begin{array}{c} 1 & 0 \\ - 1 & 1 \\ 0 & - 1 \end{array}]}_{3 \times 2} {[\begin{array}{c} u_{1} (t) \\ u_{2} (t) \end{array}]}_{2 \times 1} \\ = \underset{modeling matrix}{A} \cdot \underset{algebraic worker}{\vec{u} (t)} \end{aligned}

Recall when the algebraic worker is on the right and the modeling matrix is on the left, we work with columns.
\end{proof}

Net Spring Force Vector

Now that we have our elongation matrix, we can come up with our force vector.

\begin{proof}[Solution.]
Let $f_{s i} (t)$ be the net change in stored force of spring $i$ . Then, our force vector

\begin{aligned} {\vec{f}}_{s} & = [\begin{array}{c} f_{s_{1}} (t) \\ f_{s_{2}} (t) \\ f_{s_{3}} (t) \end{array}] \\ = [\begin{array}{c} k_{1} \cdot e_{1} (t) \\ k_{2} \cdot e_{2} (t) \\ k_{3} \cdot e_{3} (t) \end{array}] \end{aligned}

Remark. At rest, the springs are at a nonzero force. They are still able to do work. If you were to cut one of the springs, it would snap away. However, we can use this force value as our reference point for net zero.

$⧫$

Can we rewrite our force vector as a linear combination?

\begin{aligned} = [\begin{array}{c} k_{1} e_{1} (t) + 0 + 0 \\ 0 + k_{2} e_{2} (t) + 0 \\ 0 + 0 + k_{3} e_{3} (t) \end{array}] \\ = [\begin{array}{c} k_{1} e_{1} (t) \\ 0 \\ 0 \end{array}] + [\begin{array}{c} 0 \\ k_{2} e_{2} (t) \\ 0 \end{array}] + [\begin{array}{c} 0 \\ 0 \\ k_{3} e_{3} (t) \end{array}] \end{aligned}

In this case, we think of the scalars as the function $e (t)$ .

\begin{aligned} = e_{1} (t) [\begin{array}{c} k_{1} \\ 0 \\ 0 \end{array}] + e_{2} (t) [\begin{array}{c} 0 \\ k_{2} \\ 0 \end{array}] + e_{3} (t) [\begin{array}{c} 0 \\ 0 \\ k_{3} \end{array}] \\ = \underset{diagonal matrix!}{\underset{⏟}{{[\begin{array}{c} k_{1} & 0 & 0 \\ 0 & k_{2} & 0 \\ 0 & 0 & k_{3} \end{array}]}_{3 \times 3}}} {[\begin{array}{c} e_{1} (t) \\ e_{2} (t) \\ e_{3} (t) \end{array}]}_{3 \times 1} \\ ⟹ {\vec{f}}_{s} & = C \cdot \vec{e} (t) \end{aligned}

Note that we don't use $K$ because that's reserved for other types of really important problems in fields like engineering.

Why do we think of the scalars as $e (t)$ ? #question
\end{proof}

Net Force Equation

Let's find the net force $\sum F$ on each mass.

\begin{proof}[Solution.]

\begin{aligned} \sum F & = m \cdot a \\ ⟹ \sum F & = m \cdot \ddot{u} \end{aligned}

\begin{aligned} \underset{elongation vector}{\underset{⏟}{\vec{e} (t)}} & = A d (t) \\ ⟹ {\vec{f}}_{s} (t) & = C \vec{e} (t) \end{aligned}

Let's look at a free body diagram of mass 1.

\begin{aligned} y_{1} (t) & = net internal force on m_{1} \\ = - f_{s_{1}} (t) + f_{s_{2}} (t) + f_{e_{1}} (t) \end{aligned}

$f_{e_{1}} (t)$ is the external force applied to $m_{1}$ (ex. gravity, shaking, etc.)

Let's look at the free body diagram mass 2.

$$ \begin{align*} y_{2}(t) &= \text{net focus on } m_{2}\\ &= -f_{s_{2}}(t) + f_{s_{3}}(t) + f_{e_{2}}(t) \end{align*} $$ Therefore, our net force equation is $$ \begin{align*} \underset{ \text{net forces on masses} }{ \left[ \sum \vec{F}(t) \right]_{2 \times 1} } &= \begin{bmatrix} \sum F_{1}(t) \\ \sum F_{2}(t) \end{bmatrix}\\ &= \begin{bmatrix} -f_{s_{1}}(t) + f_{s_{2}}(t) + f_{e_{1}}(t) \\ -f_{s_{2}}(t) + f_{s_{3}}(t) + f_{e_{2}}(t) \end{bmatrix} \end{align*} $$ - & Note that the $\sum$ sign is used as the name of the vector.

We can break this vector into two pieces.

= \underset{internal forcces}{\underset{⏟}{[\begin{matrix} - f_{s_{1}} (t) + f_{s_{2}} (t) \\ - f_{s_{2}} (t) + f_{s_{3}} (t) \end{matrix}]}} + \underset{external forces}{\underset{⏟}{[\begin{matrix} f_{e_{1}} (t) \\ f_{e_{2}} (t) \end{matrix}]}}

What would happen if we factored out the negative?

= - [\begin{matrix} f_{s_{1}} (t) - f_{s_{2}} (t) + 0 \\ 0 + f_{s_{2}} (t) - f_{s_{3}} (t) \end{matrix}] + [\begin{matrix} f_{e_{1}} (t) \\ f_{e_{2}} (t) \end{matrix}]

Do you see a 6.1 Linear Combination of Vectors?

= - ([\begin{matrix} f_{s_{1}} (t) \\ 0 \end{matrix}] + [\begin{matrix} - f_{s_{2}} (t) \\ f_{s_{2}} (t) \end{matrix}] + [\begin{matrix} 0 \\ f_{s_{3}} (t) \end{matrix}]) + [\begin{matrix} f_{e_{1}} (t) \\ f_{e_{2}} (t) \end{matrix}]

If we pull out the scalars, we see:

= - (f_{s_{1}} (t) [\begin{matrix} 1 \\ 0 \end{matrix}] + f_{s_{2}} (t) [\begin{matrix} - 1 \\ 1 \end{matrix}] + f_{s_{3}} (t) [\begin{matrix} 0 \\ 1 \end{matrix}]) + [\begin{matrix} f_{e_{1}} (t) \\ f_{e_{2}} (t) \end{matrix}]

Using our 10.1 Matrix-Column-Vector Multiplication via Linear Combinations definition, we see

\begin{aligned} = - {[\begin{array}{c} 1 & - 1 & 0 \\ 0 & 1 & - 1 \end{array}]}_{2 \times 3} {[\begin{array}{c} f_{s_{1}} (t) \\ f_{s_{2}} (t) \\ f_{s_{3}} (t) \end{array}]}_{3 \times 1} + [\begin{array}{c} f_{e_{1}} (t) \\ f_{e_{2}} (t) \end{array}] \\ = - A^{T} \cdot {\vec{f}}_{s} (t) + {\vec{f}}_{e} (t) \end{aligned}

Notice how our matrix is the transpose of our modeling matrix for the net spring force.

\end{proof}

Matrix-Matrix Multiplication and Mass-Spring Chains

How can we use 11 Matrix-Matrix Multiplication to represent our mass-spring chain system?

\begin{proof}[Solution.]
Let $- \vec{y} (t) = A^{T} {\vec{f}}_{s} (t)$ .

- A^{T} \cdot {\vec{f}}_{s} (t) + {\vec{f}}_{e} (t) = y (t) + {\vec{f}}_{e} (t)

Recall net forces on masses:

\begin{aligned} [\begin{array}{c} \sum F_{1} (t) \\ \sum F_{2} (t) \end{array}] = {[\sum \vec{F} (t)]}_{2 \times 1} & = \vec{y} (t) + {\vec{f}}_{e} (t) \\ = - A^{T} \cdot {\vec{f}}_{s} (t) + {\vec{f}}_{e} (t) \\ = - A^{T} \cdot C \cdot \vec{e} (t) + {\vec{f}}_{e} (t) \\ = \underset{\underset{stiffness matrix}{K_{2 \times 2}}}{\underset{⏟}{[- A^{T}]_{2 \times 3} \cdot [C]_{3 \times 3} \cdot [A]_{3 \times 2}}} \cdot [\vec{u} (t)]_{2 \times 1} + [\vec{f} e (t)]_{2 \times 1} \\ ⟹ {[\sum \vec{F} (t)]}_{2 \times 1} & = - K \cdot \vec{u} (t) + {\vec{f}}_{e} (t) \\ ⟹ [\begin{array}{c} \sum F_{1} (t) \\ \sum F_{2} (t) \end{array}] & = [\begin{array}{c} m_{1} \cdot \ddot{u_{1}} (t) \\ m_{2} \cdot \ddot{u_{2}} (t) \end{array}] \end{aligned}

Can this be a linear combination?

\begin{aligned} = [\begin{array}{c} m_{1} \cdot \ddot{u_{1}} + 0 \\ 0 + m_{2} \cdot \ddot{u_{2}} \end{array}] \\ [\begin{array}{c} \sum F_{1} (t) \\ \sum F_{2} (t) \end{array}] & = [\begin{array}{c} m_{1} \cdot \ddot{u_{1}} \\ 0 \end{array}] + [\begin{array}{c} 0 \\ m_{2} \cdot \ddot{u_{2}} \end{array}] \\ = \ddot{u_{1}} (t) [\begin{array}{c} m_{1} \\ 0 \end{array}] + \ddot{u_{2}} [\begin{array}{c} 0 \\ m_{2} \end{array}] \\ = \underset{M_{2 \times 2}}{\underset{⏟}{{[\begin{array}{c} m_{1} & 0 \\ 0 & m_{2} \end{array}]}_{2 \times 2}}} \underset{[\ddot{\vec{u}} (t)]_{2 \times 1}}{\underset{⏟}{{[\begin{array}{c} \ddot{u_{1}} (t) \\ \ddot{u_{2}} (t) \end{array}]}_{2 \times 1}}} \\ ⟹ M \ddot{\vec{u}} (t) & = - K \vec{u} (t) + {\vec{f}}_{e} (t) \\ ⟹ M \cdot \ddot{\vec{u}} (t) + K \vec{u} (t) & = {\vec{f}}_{e} (t) \end{aligned}

If we think back to when gravity is turned on,

\ddot{\vec{u}} (t) = [\begin{matrix} 0 \\ 0 \end{matrix}] since masses are at rest and {\vec{f}}_{e} (t) = [\begin{matrix} 9.8 \cdot m_{1} \\ 9.8 \cdot m_{2} \end{matrix}]

Therefore, we can represent our system as

\begin{aligned} ⟹ M \cdot {\ddot{\vec{u}} (t)}^{\vec{0}} + K \vec{u} (t) & = {\vec{f}}_{e} (t) \\ ⟹ K \vec{u} (t) & = {\vec{f}}_{e} (t) \\ ⟹ K \cdot \vec{u} & = {\vec{f}}_{e} \end{aligned}

\end{proof}