Lesson 13 - Mass Spring Systems

This lesson will bring what we've learned about nonhomogeneous differential equations into a real world application.

Consider a mass $m$ that is attached to the end of a spring with spring constant $k$ . The mass is stretched from the equilibrium position by $x (t)$ . The force due to damping $F_{d}$ is proportional to the velocity of the mass, that is, $F_{d} = - b x^{'}$ .

Hooke's Law states that the force excreted by the spring on the mass is $F_{s} = - k x .$

If $F_{e x t} (t)$ is an external force on the mass then applying Newton's second law of motion gives us the differential equation modeling the mass-spring oscillator, and we can derive a differential equation from these relations.

(1) m a = F_{e x t} - F_{s} - F_{d}

(2) m x^{″} = F - k_{x} - b x^{'}

(3) m x^{″} + b x^{'} + k x = F_{e x t}

where $m$ is the mass of the object, $b$ is the damping force, and $k$ is the distance that the spring has moved from equilibrium.

Now, we have a nonhomogeneous equation that we have seen before. Also consider that $m, b, k$ are all constants that are greater than $0$ .

Since we are describing an oscillating force, it will be helpful for us to see the behavior of the oscillating graph. We can do so by solving the differential equation and then converting the general solution into a sine form.

For the above model, write the general solution when the damping force is zero and there is no external force. Let $ω = \sqrt{\frac{k}{m}} .$ Show that the solution can be written as $x (t) = A \sin (ω t + ϕ)$ for a suitable $A, ϕ$ .

We start with the equation of motion with no external force and damping force of zero: $m x^{″} + k x = 0$ .

We can now use the characteristic equation to find the general solution.

m x^{″} + k x = 0

m r^{2} + k = 0

r = \pm i \sqrt{\frac{k}{m}}

Substituting $ω = \sqrt{\frac{k}{m}}$

r = i ω

Recall the general solution of nonhomogeneous equations using Euler's formula (see Lesson 9 - Homogeneous Second Order Linear Equations). The general solution is as follows:

x (t) = C_{1} \cos (ω t) + C_{2} \sin (ω t)

Now, consider a right triangle with sides $C_{1}$ and $C_{2}$ .

\usepackage{tikz}
\begin{document}
\begin{tikzpicture}
  \draw[thick] (0, 0) -- (3, 0) -- (3, 4) -- cycle; % Triangle
  \draw (3, 0) rectangle +(-0.5, 0.5); % Right angle marker
  \draw[thick] (0, 0) -- (3, 4) node[left, midway] {$\sqrt{(C_1)^2+(C_2)^2}$}; % Hypotenuse label
  \draw[thick] (0, 0) -- (3, 0) node[below, midway] {$C_2$}; % Hypotenuse label
  \draw[thick] (3, 0) -- (3, 4) node[right, midway] {$C_1$}; % Hypotenuse label
  \draw[thick] (0, 0) -- (0.35, 0) arc[start angle=0, end angle=60, radius=0.3cm]; % Angle arc 
  \node at (0.5, 0.2) {$\phi$}; % Label the angle
\end{tikzpicture}
\end{document}

Imposing geometry, by the Pythagorean theorem, we can say that the sides $(C_{1})^{2}$ and $(C_{2})^{2}$ give us a hypotenuse $\sqrt{(C_{1})^{2} + (C_{2})^{2}}$ . Let's call this $A$ .

$A = \sqrt{(C_{1})^{2} + (C_{2})^{2}}$

We also impose an angle $ϕ$ . From this, we can set up the following relations:

$\sin ϕ = \frac{C_{1}}{\sqrt{(C_{1})^{2} + (C_{2})^{2}}}$
$\cos ϕ = \frac{C_{2}}{\sqrt{(C_{1})^{2} + (C_{2})^{2}}}$

Now let's manipulate the general solution equation to contain $A$ . We do so by multiplying the equation by $\frac{A}{A}$ or $\frac{\sqrt{(C_{1})^{2} + (C_{2})^{2}}}{\sqrt{(C_{1})^{2} + (C_{2})^{2}}}$ . Essentially, we are multiplying the equation by 1, which does not alter the equation, but allows us to manipulate it to get the form that we want.

\frac{\sqrt{(C_{1})^{2} + (C_{2})^{2}}}{\sqrt{(C_{1})^{2} + (C_{2})^{2}}} [C_{1} \cos ω t + C_{2} \sin ω t]

We will multiply across the denominator but leave the numerator un-factored.

\sqrt{(C_{1})^{2} + (C_{2})^{2}} [\frac{C_{1}}{\sqrt{(C_{1})^{2} + (C_{2})^{2}}} \cos ω t + \frac{C_{2}}{\sqrt{(C_{1})^{2} + (C_{2})^{2}}} \sin ω t]

Recall that $A = \sqrt{(C_{1})^{2} + (C_{2})^{2}}$ and our trig relations that we set up before. We can make some substitutions.

x (t) = A [\sin ϕ \cos ω t + \cos ϕ \sin ω t]

Notice we have a sum of the products of two angles. We can implement the following identity:

\sin (θ \pm ϕ) = \sin θ \cos ϕ \pm \cos θ \sin ϕ

The forms look different, but they're the same if we rearrange the equation and consider $ω t = θ$ .

x (t) = A [\sin ω t \cos ϕ + \cos ω t \sin ϕ]

Now these are identical forms and it's more clear that we can use this identity. Our general solution is now as follows:

x (t) = A \sin (ω t + ϕ)

So why did we do all of this work if the past solutions have worked in the past? Well consider that we are now talking about sine functions as we analyze the oscillations of the the mass spring system. This form helps us to analyze the angular frequency ( $ω$ ), the natural frequency ( $\frac{ω}{2 π}$ ), the period ( $\frac{2 π}{ω}$ ), and the amplitude in a clear and concise manner.

We can see an example of this below:

\usepackage{pgfplots}
\begin{document}
\begin{tikzpicture}
\begin{axis}[
    axis lines = middle,
    xlabel = {$t$},
    ylabel = {$x(t)$},
    grid = both,
    domain=-12:12*pi,
    samples=100,
    legend style={at={(0.5,-0.15)}, anchor=north},
    title = {Oscillation: $x(t) = A \sin(\omega t + \phi)$},
    xtick=\empty,
    ytick=\empty,
]
% General sine function with parameters A, omega, and phi
\addplot[blue, thick] {sin(deg(x))}; 
\addlegendentry{$x(t) = \sin(t)$}

\end{axis}
\end{tikzpicture}
\end{document}

As we can see here, it is much easier to see the behavior of the oscillation system by converting the general solution into it's a sinusoidal form.

In the next mini lesson, we will look at a couple of examples of this and discuss free vibrations with damping.

Next Lesson: Lesson 13.1 Mass Spring Systems Damping and Worked Examples

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