Math 555: Differential Equations

3.7 Mechanical and Electrical Vibrations

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3.7.1 Modeling with Second Order Equations

Second-order linear equations with constant coefficients are useful because they provide a model for the behavior of many physical systems. In particular, they describe the behavior of systems involving a mechanical or electrical oscillation. A mass hanging on a spring, the torsional oscillation of a shaft with a flywheel, and the way a current flows through a circuit in series may all be modeled using the initial value problem

$$ ay'' + by' + cy = g(t),\quad y(0) = y_0,\quad y'(0) = y_0'. $$
We have developed our understanding of how to solve these sorts of IVPs and will now apply that understanding to modeling systems we can describe using this second-order equation. Notably, there are no new mathematics we need to learn here. All of the physical processes we will study in this section will simply be applications of techniques we are familiar with. The new elements will be assigning physical significance to the constants $a$, $b$, and $c$; the independent variable $t$ (typically time); and the dependent variable $y$ (though we may change the letter used to make the physical meaning more clear).

This is a really important connection between the realms of mathematics and physics or other sciences. The mathematics underlying these models is consistent regardless of the physical interpretation. "Filing the names off," so to speak, provides the opportunity to study the underlying mathematical principles guiding how a system functions and may help you see connections in places you may not otherwise expect it. Keep your eyes open and try to develop an intuition for how different systems in various settings such as mechanics, electronics, finance, chemistry, etc. share ideas. There are many similarities to be found and keen insights to be made. Finding these will greatly sharpen your critical thinking skills and make you a better problem solver.

3.7.2 Mechanical Vibrations: Hanging Mass

Now we focus on the specific application of mechanical vibrations, specifically those of a mass $m$ hanging on a spring. The spring is stretched an amount $L$ from its natural length, and because the system is at rest, the sum of the forces must be equal to zero. Our goal is to develop a model for the motion in this hanging mass system in the event that the mass is moved vertically away from this position and released. Before we begin, we need to make a few assumptions.

1. The spring is not damaged by the hanging mass.
   In real life situations, it is possible to hang too heavy a mass and mechanically fatigue the spring, changing its physical properties. We are assuming that does not happen here.
2. The spring is not stretched or compressed more than physically possible.
   Similar to the first condition, there are bounds to how much the spring can be compressed or stretched before physical limits come into play. For example, stretching the spring too far will cause the sort of mechanical fatigue mentioned in assumption 1. Also, when the spring is compressed, the coils must get closer together. At some point, the would need start to touch and spring can not be further compressed without damaging it. We assume neither of these situations arise.
3. All motion is completely vertical.
   This model is only for vertical displacement. We assume that the mass does not swing in the horizontal direction.

   Pendulums are nonlinear ODEs. We're not going to mess with that at the moment.

4. The initial conditions of the system do not violate any of the previous assumptions.
   The easiest way to break the above rules is to start the system with a very large displacement or velocity for the mass. It is easy to imagine pulling down way too far on the mass and letting go, causing it to snap back wildly. Our model isn't going to cover that case.

Setting Up the Model

We begin with one of the two most famous formulas in physics:

Newton's Second Law

$$\text{force} = \text{mass}\cdot\text{acceleration}.$$

In this instance, we will be using this equation to look at the force being exerted by the mass on the spring due to weight

$$ F_w = mg. $$
We know that in a resting state, the sum of the forces must be zero, so there is a force that is balancing out the weight of the mass. This force is coming from the spring and is described by Hooke's law

$$ F_s = -kL, $$
where $L$ is the displacement of the spring from its natural length and $k$ is a constant of proportionality specific to the spring, called the spring constant . (One of our assumptions is that this constant stays the same, as mechanical fatigue would cause this value to change.)

Of particular importance to Hooke's law is the fact that the force is opposite the direction of displacement. The spring resists being stretched (or compressed) and wishes to return to its natural length, pulling (or pushing) in order to achieve that. The two forces here are in balance, so

$$ F_w + F_s = mg - kL = 0, $$
and we have the associated free body diagram:

To build the model, we are going to define a function $u(t)$ which represents the displacement away from the rest position for the hanging mass at time $t$, where $u(t) > 0$ represents that the spring is further stretched and $u(t) < 0$ is a compression. The model takes into account all of the forces acting on the system, using Newton's second law with the acceleration $u''(t)$ (since this is the second derivative of the displacement function $u(t)$),

$$ mu''(t) = f(t) $$
where $f(t)$ is the net force on the system. The net force is composed of the following components:

1. The weight of the hanging mass $F_w = mg$.
2. The force of the spring trying to return its equilibrium position $F_s = -k(L + u(t))$.
3. The force of viscous damping . This force is proportional to the velocity of the mass, and represents the mass moving through a fluid. It is a type of drag in the aero- or fluid dynamic sense, and appears in our model as $F_d = -\gamma u'(t)$ since it opposite the direction of motion.

   There are many models of types of drag in physical systems, and this is one of the simplest. These sort of resistive forces are almost always nonlinear in nature, so we just employ an approximation here.

4. Applied external forces $F(t)$. Previously in discussions this chapter, the term forcing function was used to describe the nonhomogeneous part of our second order equations. This is because in applications, nonhomogeneous equations are the result of an applied external force to the system. We will see that the natural tendency of our system here is a simple oscillation, but applying an external force will alter this behavior.

Adding all of these terms

$$ \begin{aligned} mu''(t) &= F_w + F_s + F_d + F(t) \\ &= mg - k(L+u(t)) - \gamma u'(t) + F(t) \\ &= \underbrace{mg - kL}_0 - ku(t) - \gamma u'(t) + F(t), \end{aligned} $$
we can simplify the equation into the form for a second-order linear nonhomogeneous equation with constant coefficients

$$ mu''(t) + \gamma u'(t) + ku(t) = F(t) $$
for positive constants $m$, $\gamma$, and $k$.

Damping and Its Effect on Solutions

Damped oscillations were introduced in Section 3.4 . Here, we utilize the ideas introduced to analyze the applications of mechanical vibrations.

Undamped Systems

If $\gamma = 0$, then the characteristic equation of our ODE is $ mr^2 + k = 0$. Find the roots, we have that

$$ r = \pm\sqrt{-\frac{k}{m}} = \pm i\sqrt{\frac{k}{m}}$$
and so our solutions take the form

$$ \begin{aligned} u(t) &= A\cos\sqrt{\frac{k}{m}}t + B\sin \sqrt{\frac{k}{m}}t \\ \\ &= A\cos\omega_0 t + B\sin\omega_0 t \end{aligned} $$
where the quantity $\omega_0 = \sqrt{\frac{k}{m}}$, measured in radians per unit time, is referred to as the natural frequency of the system. It is related to the period $T$ of the oscillation, which is the amount of time required for one complete wave form

$$ T = \dfrac{2\pi}{\omega_0}. $$
For any expression in the form $A\cos\omega_0 t + B\sin\omega_0 t$, it is possible to rewrite it in the form $R\cos\left(\omega_0 t - \delta\right)$ for

$$ R = \sqrt{A^2 + B^2},\quad \tan\delta = \frac{B}{A}. $$
This allows us to more easily determine the maximum displacement from equilibrium or amplitude $R$ and horizontal shift or phase angle $\delta$ of the system. Please note that care must be taken with determining the phase, since the inverse tangent function only has a range of $[-\frac{\pi}{2},\frac{\pi}{2}]$. The phase angle must be adjusted to be in the correct quadrant. My favored approaches for this are to use reference angles, since $|\arctan(t)|$ is always equal to the reference angle, or to use a numerical function atan2 that is found in many scientific computing software packages.

3.7.3 Undamped Mechanical Vibration

Example 3.7.1

A mass weighing 3 lbs stretches a spring 3 in. If the mass is pushed upward, contracting the spring a distance of 1 in and then set in motion with a downward velocity of 2 ft/s, and if there is no damping, find the position $u$ of the mass at any time $t$. Determine the frequency, period, amplitude, and phase of the motion.

Solution:

We take the given information and convert place it in our model

$$ mu''(t) + \gamma u'(t) + ku(t) = F(t) $$
where

$$ m = \dfrac{F_w}{g} = \dfrac{3 \text{ lbs}}{32 \text{ ft/s}^2} = \dfrac{3}{32} \dfrac{ \text{lbs}\cdot \text{s}^2}{\text{ft}}, $$
$\gamma = 0$ due to no damping, and $k$ is found by using Hooke's law

$$ k = \dfrac{F_s}{L} = \dfrac{3 \text{ lbs}}{\frac{3}{12} \text{ ft}} = 12 \text{ lbs/ft}. $$
In this problem, units are very important. It is necessary to make sure that the proper conversions are made so that each quantity is assigned the correct value and that units are consistent. Here, I have converted all distance measurements to feet. Inches are also valid, but would need to be used in all parts of the problem. Also, the force measurement given in lbs was converted to mass for use in the equation. Please do not forget to do this in problems using imperial units .

The differential equation is

$$ \dfrac{3}{32}u''(t) + 12u(t) = 0, $$
with initial values

$$ u(0) = -\dfrac{1}{12} \text{ ft},\quad u'(0) = 2\text{ ft/s}. $$

Since $\gamma = 0$, this is an example of undamped oscillations . The frequency of the oscillations is

$$ \omega_0 = \sqrt{\dfrac{k}{m}} = 8\sqrt{2}\text{ rad/s}, $$
which means the period is

$$ T = \dfrac{\pi\sqrt{2}}{8} \text{ s}. $$
The solution to the initial value problem is

$$ u(t) = -\dfrac{1}{12}\cos 8\sqrt{2}t + \dfrac{\sqrt{2}}{8}\sin 8\sqrt{2} t. $$

We consider the equivalent expression $R\cos\left(\omega_0 t - \delta\right)$ with

$$ R = \sqrt{A^2 + B^2},\quad \tan\delta = \frac{B}{A}. $$
Here the amplitude is given by

$$ R = \sqrt{\frac{1}{144} + \frac{2}{64}} = \sqrt{\frac{11}{288}}\approx 0.1954\text{ ft},$$
and for the phase $\delta$ we notice that since $A = -\frac{1}{12} \lt 0$ and $B = \frac{\sqrt{2}}{8} \gt 0$ the phase $\delta$ must be in the second quadrant, so

$$ \delta = \pi - \left\vert \arctan\left(\frac{B}{A}\right)\right\vert = \pi - \arctan\left(\frac{3\sqrt{2}}{2}\right) \approx 2.0113\text{ rad}.$$

3.7.3 Damped Mechanical Vibrations

In the case where the damping coefficient $\gamma$ in nonzero, the solutions are not freely oscillating as in the previous example. The solution to the equation

$$ mu''(t) + \gamma u'(t) + ku(t) = 0 $$
has three cases, where the roots of the characteristic equation $mr^2 + \gamma r + k = 0$ are

$$ r_1,r_2 = \dfrac{\gamma \pm \sqrt{\gamma^2 - 4km}}{2m}. $$
The types of conditions that arise are discussed in Section 3.4 , based on the sign of the discriminant $\gamma^2 - 4km$

1. Overdamped:
   $\gamma^2 - 4km \gt 0 \Rightarrow$
   
   $$ u(t) = Ae^{r_1 t} + Be^{r_2 t} $$
   
2. Underdamped:
   $\gamma^2 - 4km \lt 0 \Rightarrow$
   $$ u(t) = e^{-\frac{\gamma t}{2m}}\left(A\cos(\mu t) + B\sin(\mu t)\right),\quad \mu = \frac{1}{2m}\sqrt{4km - \gamma^2} $$
   
3. Critically Damped :
   $\gamma^2 - 4km = 0 \Rightarrow$
   
   $$ u(t) = (A + Bt)e^{-\frac{\gamma t}{2m}}. $$

Example 3.7.2

A mass of 20 g stretches a spring 5 cm. Suppose that the mass is also attached to a viscous damper with a damping constant of 400 dyn·s/cm. If the mass is pulled down an additional 2 cm and then released, find its position $u$ at any time $t$. Plot $u$ versus $t$. Determine the quasi-frequency and the quasi-period. Determine the ratio of the quasi-period to the period of the corresponding undamped motion. Also find the time $\tau$ such that $|u(t)| \lt 0.05$ cm for all $t \gt \tau$.

Solution:

First, dynes are a unit of force given by

$$ \text{dyn} = \dfrac{\text{g}\cdot\text{cm}}{\text{s}^{2}}. $$
This problem will be worked using cgs units, since that is how the quantities were given. The mass $m$ is 20 g. The damping constant $\gamma$ is 400 dyn·s/cm, and the spring constant

$$ k = \dfrac{mg}{L} = \dfrac{20\text{ g}\cdot 980 \text{ cm/s}^{2}}{5 \text{ cm}} = 3920 \text{ dyn/cm}. $$
Our differential equation is then

$$ 20u'' + 400u' + 3920u = 0, $$
which simplifies to

$$ u'' + 20u' + 196 = 0 $$
with initial conditions

$$ u(0) = 2,\quad u'(0) = 0. $$
Since $\gamma^2 - 4km = (20)^2 - 4(196)(1) \lt 0$, this system is underdamped , and the roots of the characteristic equation are

$$r = \dfrac{-20 \pm \sqrt{(20)^2 - 4(196)(1)}}{2(1)} = -10 \pm 4\sqrt{6}i. $$
The general solution is

$$ u(t) = e^{-10t}\left(A\cos(4\sqrt{6} t) + B\sin(4\sqrt{6} t)\right). $$
Applying the initial conditions, we may solve for $A$ and $B$ to determine that the solution to the initial value problem is

$$ u(t) = e^{-10t}\left(2\cos(4\sqrt{6} t) + \dfrac{5\sqrt{6}}{6}\sin(4\sqrt{6} t)\right). $$

We have been asked to find the quasi-frequency $\mu = 4\sqrt{6}$, the quasi-period

$$ T_d = \frac{2\pi}{\mu} = \frac{2\pi}{4\sqrt{6}} = \frac{\pi\sqrt{6}}{12}, $$
and the ratio of the quasi-period to the period of the undamped system

$$ u'' + 196u = 0 $$
whose natural frequency is $w_0 = 14$ rad/s. Therefore the ratio of the quasi-period to the undamped period is

$$ \frac{T_d}{T} = \dfrac{\frac{\pi\sqrt{6}}{12}}{\frac{2\pi}{14}} = \frac{7\sqrt{6}}{12}\approx 1.4289. $$
Lastly, we find the time $\tau$ where $t \gt \tau\Rightarrow |u(t)| \lt 0.05$. The inequality

$$ \left\vert e^{-10t}\left(2\cos(4\sqrt{6} t) + \dfrac{5\sqrt{6}}{6}\sin(4\sqrt{6} t)\right) \right\vert < 0.05 $$
is very difficult to solve, so we employ computer software (I used MATLAB ) to determine that $\tau\approx 0.4045$ s.

The figure here shows $u(t)$ versus $t$ and the bounds $\pm 0.05$.

3.7.4 Electrical Vibrations: RLC Series Circuit

Another system that can be modeled with a second-order linear equation with constant coefficients is a RLC circuit in series. This type of circuit is a single loop where $R$ is the resistance in ohms $\Omega$, $L$ is an inductance in henrys H, $C$ is a capacitance in farads F, and $E(t)$ is the impressed voltage in volts V. (All of these units are named after people, hence the capital letters for symbols.)

In this model, we combine several principles from basic circuits. The second-order equation we are about to derive will be a measure of the total charge $Q$ in coulombs C on the capacitor as a function of time $t$. We begin with the relationship between charge and current (change or flow of charge)

$$ I = \dfrac{dQ}{dt}. $$
The current flow through the circuit is described by Kirchhoff's voltage law:

Kirchhoff's Voltage Law

The directed sum of the potential differences (voltages) around any closed loop is zero.

This means that if we follow a path around the circuit, the total of the voltage drops over the $R$, $L$, and $C$ components must be equal to the impressed voltage $E(t)$. These drops are

• $IR$ for the resistor, by Ohm's law .
• $\frac{Q}{C}$ for the capacitor, since capacitance is defined as $C = \frac{Q}{V}$.
• $L\frac{dI}{dt}$ for the inductor.

These may be combined into the expression

$$ L\frac{dI}{dt} + IR + \frac{1}{C}Q = E(t), $$
which may be rewritten after we remember that $I = Q'(t)$ to form a second-order linear ODE with constant coefficients:

$$ LQ''(t) + RQ'(t) + \dfrac{1}{C}Q(t) = E(t). $$

Example 3.7.3

A series circuit has a capacitor of $10^{-5}$ F, a resistor of $3\cdot 10^2\ \Omega$, and an inductor of $0.2$ H. The initial charge on the capacitor is $10^{-6}$ C and there is no initial current. Find the charge $Q$ on the capacitor at any time $t$. Additionally, determine what the resistance $R$ would need to be for this system to be critically damped.

Solution:

Our initial problem is

$$ \left(0.2\right) Q'' + \left(3\cdot 10^2\right)Q' + \left(10^{5}\right)Q = 0,\quad Q(0) = 10^{-6},\ Q'(0) = 0. $$
It has the characteristic equation

$$ 0.2r^2 + 300r + 100000 = 0\quad\text{or}\quad r^2 + 1500r + 500000 = 0 $$
with roots

$$ r_1 = -500\qquad r_2 = -1000. $$
This means that the general solution is

$$ Q(t) = Ae^{-500t} + Be^{-1000t}. $$
Applying the initial conditions $Q(0) = 10^{-6}$ and $Q'(0) = 0$,

$$ \begin{aligned} Q(0) = 10^{-6}\quad &\Rightarrow \qquad\quad\ A + \qquad\!\! B = 10^{-6} \\ Q'(0) = 0\qquad &\Rightarrow \quad -500A - 1000B = 0. \end{aligned} $$
This is a $2\times 2$ system whose solution is $A = 2\cdot 10^{-6}$, $B = -10^{-6}$, and so the solution to the initial value problem is

$$ Q(t) = 2\cdot 10^{-6}e^{-500t} - 10^{-6}e^{-1000t}. $$
For this system to be critically damped, we require that the discriminant of the characteristic equation

$$ Lr^2 + Rr + \dfrac{1}{C}r = 0 $$ be equal to zero, this means that for a critically damped system,

$$ R^2 - 4\frac{L}{C} = 0 \quad\Rightarrow\quad R = 2\sqrt{\frac{L}{C}}. $$
In this case, the resistance to critically damp the system must be

$$ R = 2\sqrt{\frac{0.2}{10^{-6}}} = 200\sqrt{2} \ \Omega \approx 282.8\ \Omega. $$
With the original resistance of $300\ \Omega$, the system is overdamped. Here is a figure comparing the two systems: We see that since $R$ in the original problem is fairly close the critical value, these curves are quite similar.

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