← Back to Math 055

5.1 Linear Models: Initial-Value Problems: Class Notes

5.1a Mass-Spring Models Setup

5.1b

Undamped motion

Example

A mass weighing 32 pounds stretches a spring 2 feet. Determine the amplitude and period of motion if the mass is initially released from a point 1 foot above the equilibrium position with an upward velocity of 2 ft/s. How many complete cycles will the mass have completed at the end of $4\pi$ seconds?

Spring/mass system in three stages (after Figure 5.1.1 in the textbook). (a) An unstretched spring of natural length $l$ hangs from a rigid support. (b) A mass $m$ is attached and the spring stretches an extra $s$ to the equilibrium position, where $mg-ks=0$. (c) In motion, the mass is a distance $x$ beyond the equilibrium position, so the spring is stretched $s+x$ beyond its natural length (the equilibrium position sits $l+s$ below the support).

Solution

Setting up the model

Hooke’s Law: $F=k|s|$. The weight is the force and the stretch is $F=32\text{ lb},|s|=2\text{ ft}$, so

$$32=k·2⟹k=16\text{ lb/ft}$$

At the equilibrium position (b), the force due to gravity, $mg$, balances the restoring force of the spring, $ks$:

$$mg=ks⟹mg-ks=0$$

Assuming down is positive, if the mass is moving, we have $F=ma$ (Newton’s 2nd law) with ${\displaystyle a=\frac{d^{2}x}{d{t}^2}}$. So

$${\displaystyle \begin{array}{cc}\hfill {\displaystyle m\frac{d^{2}x}{d{t}^2}}& {\displaystyle =-k(x+s)+mg}\hfill \\ \hfill {\displaystyle m\frac{d^{2}x}{d{t}^2}}& {\displaystyle =-kx\underset{=0}{\underbrace{-ks+mg}}}\hfill \end{array}}$$

The last two terms are zero by the equilibrium condition, which leaves

$${\displaystyle m\frac{d^{2}x}{d{t}^2}+kx=0}$$

where $x(t)$ is distance from equilibrium, $m$ is the mass, and $k$ is the spring constant.

Units

In the second row, the mass in slugs is the weight in pounds divided by 32.

Solving

The mass weighs 32 lb. Weight is $mg$, with $g=32{\text{ ft/s}}^2$:

$$32=32m⟹m=1$$

Dividing ${\displaystyle m\frac{d^{2}x}{d{t}^2}+kx=0}$ by $m$ and letting ${\displaystyle {\omega }^2=\frac{k}{m}}$ yields

$${\displaystyle \frac{d^{2}x}{d{t}^2}+{\omega }^{2}x=0⟹r^2+{\omega }^2=0\text{(auxiliary equation)}}$$

so $r=\pm \omega i$, where ${\displaystyle \omega =\sqrt{\frac{k}{m}}}$. This gives

$$x(t)=c_1\cos \omega t+c_2\sin \omega t$$

For the above example, $k=16$ and $m=1$, so ${\displaystyle \omega =\sqrt{\frac{16}{1}}=4}$ and

$$x(t)=c_1\cos 4t+c_2\sin 4t$$

Released from 1 ft above equilibrium (up is negative): $x(0)=-1$

Upward velocity 2 ft/s: $x\prime (0)=-2$

$x(0)=-1$ gives $c_1=-1$. Next, $x\prime (t)=-4c_1\sin 4t+4c_2\cos 4t$, so

$${\displaystyle x\prime (0)=-2⟹-2=4c_2⟹c_2=-\frac{1}{2}}$$

Equation of motion:

$${\displaystyle x(t)=-\cos 4t-\frac{1}{2}\sin 4t}$$

Period of motion:

$${\displaystyle T=\frac{2\pi }{\omega }=\frac{2\pi }{4}=\frac{\pi }{2}\text{ sec/cycle}}$$

Frequency:

$${\displaystyle f=\frac{1}{T}=\frac{2}{\pi }\text{ cycles/sec (hertz)}}$$

At the end of $4\pi$ seconds, the number of cycles is ${\displaystyle \frac{4\pi }{\pi /2}=8}$.

To find the amplitude

Consider $a\cos \theta +b\sin \theta$. Factor out $\sqrt{a^2+b^2}$:

$${\displaystyle \sqrt{a^2+b^2}(\frac{a}{\sqrt{a^2+b^2}}\cos \theta +\frac{b}{\sqrt{a^2+b^2}}\sin \theta )}$$

Right triangle with acute angle $\phi$. Opposite side: $a$. Adjacent side: $b$. Hypotenuse: $\sqrt{a^2+b^2}$.

Suppose there is an angle $\phi$ such that ${\displaystyle \frac{a}{\sqrt{a^2+b^2}}=\sin \phi }$ and ${\displaystyle \frac{b}{\sqrt{a^2+b^2}}=\cos \phi }$ (the triangle above, with $a$ opposite $\phi$ and $b$ adjacent).

Then we have

$$\begin{array}{cc}\hfill {\displaystyle \sqrt{a^2+b^2}(\sin \phi \cos \theta +\cos \phi \sin \theta )}& {\displaystyle =\sqrt{a^2+b^2}\sin (\theta +\phi )}\hfill \end{array}$$

So if $x(t)=c_1\cos \omega t+c_2\sin \omega t$, then

$${\displaystyle x(t)=\sqrt{c_1^2+c_2^2}\sin (\omega t+\phi ),\text{where }\tan \phi =\frac{c_1}{c_2}}$$

For our example, ${\displaystyle x(t)=-\cos 4t-\frac{1}{2}\sin 4t}$, so

$${\displaystyle \sqrt{c_1^2+c_2^2}=\sqrt{1+\frac{1}{4}}=\frac{\sqrt{5}}{2}}$$

$${\displaystyle \tan \phi =\frac{c_1}{c_2}=2⟹{\phi }_{\text{ref}}\approx 1.11}$$

But $\sin \phi$ and $\cos \phi$ have the signs of $c_1=-1$ and ${\displaystyle c_2=-\frac{1}{2}}$, both negative, so $\phi$ is in Quadrant III:

$$\phi =\pi +1.11\approx 4.25$$

$${\displaystyle x(t)=\frac{\sqrt{5}}{2}\sin (4t+4.25)}$$

Generalization

${\displaystyle m\frac{d^{2}x}{d{t}^2}+kx=0}$ yields

$$x(t)=c_1\cos \omega t+c_2\sin \omega t$$

$${\displaystyle x(t)=\sqrt{c_1^2+c_2^2}\sin (\omega t+\phi ),\text{where }\tan \phi =\frac{c_1}{c_2}}$$

Right triangle with acute angle $\phi$. Opposite side: $c_1$. Adjacent side: $c_2$. Hypotenuse: $\sqrt{c_1^2+c_2^2}$.

$F=k|s|$ gives $k$ ($F$ is force: lb, newtons), $m$ is mass (kg, slugs), and ${\displaystyle \omega =\sqrt{\frac{k}{m}}}$.

It is also possible to express the solution in terms of cosine:

$${\displaystyle m\frac{d^{2}x}{d{t}^2}+kx=0⟹x(t)=c_1\cos \omega t+c_2\sin \omega t}$$

$${\displaystyle x(t)=\sqrt{c_1^2+c_2^2}\cos (\omega t-\phi ),\text{where }\tan \phi =\frac{c_2}{c_1}}$$

Justification:

$$\cos (\alpha -\beta )=\cos \alpha \cos \beta +\sin \alpha \sin \beta$$

With $\alpha =\omega t$ and $\beta =\phi$, the cosine term matches $c_1$ and the sine term matches $c_2$, so $c_1=\sqrt{c_1^2+c_2^2}\cos \phi$ and $c_2=\sqrt{c_1^2+c_2^2}\sin \phi$, which gives ${\displaystyle \tan \phi =\frac{c_2}{c_1}}$.

Right triangle with acute angle $\phi$. Opposite side: $c_2$. Adjacent side: $c_1$. Hypotenuse: $\sqrt{c_1^2+c_2^2}$.

5.1c

Multiple springs in parallel: $k_{\text{eff}}=k_1+k_2$

Multiple springs in series: ${\displaystyle k_{\text{eff}}=\frac{k_1{k}_2}{k_1+k_2}}$

Left (parallel): two springs, with constants $k_1$ and $k_2$, side by side, each attached to both walls. Right (series): the two springs joined end to end, one after the other, between the walls.

Damped motion

$${\displaystyle m\frac{d^{2}x}{d{t}^2}=-kx-\beta \frac{dx}{dt}}$$

The left side is $ma$. On the right, $-kx$ is the restoring force from the spring, and ${\displaystyle -\beta \frac{dx}{dt}}$ is the damping force; $\beta$ is the damping constant. The damping force is proportional to the instantaneous velocity.

Dividing by $m$:

$${\displaystyle \frac{d^{2}x}{d{t}^2}+\frac{\beta }{m}\frac{dx}{dt}+\frac{k}{m}x=0}$$

Let ${\displaystyle 2\lambda =\frac{\beta }{m}}$ and ${\displaystyle {\omega }^2=\frac{k}{m}}$:

$${\displaystyle \frac{d^{2}x}{d{t}^2}+2\lambda \frac{dx}{dt}+{\omega }^{2}x=0}$$

Auxiliary equation:

$${\displaystyle \begin{array}{cc}\hfill {\displaystyle r}& {\displaystyle =\frac{-2\lambda \pm \sqrt{4{\lambda }^2-4{\omega }^2}}{2}}\hfill \\ \hfill {\displaystyle r}& {\displaystyle =-\lambda \pm \sqrt{{\lambda }^2-{\omega }^2}}\hfill \end{array}}$$

Case I: ${\lambda }^2-{\omega }^2>0$ (real, distinct roots)

$$x(t)=c_1{e}^{(-\lambda +\sqrt{{\lambda }^2-{\omega }^2})t}+c_2{e}^{(-\lambda -\sqrt{{\lambda }^2-{\omega }^2})t}$$

This does not oscillate. This system is overdamped.

Case II: ${\lambda }^2-{\omega }^2=0$

$$x(t)=c_1{e}^{-\lambda t}+c_{2}t{e}^{-\lambda t}$$

This is critically damped.

Tip: Critical damping is not going to happen by accident: it needs ${\lambda }^2={\omega }^2$ exactly.

In Case I and Case II, equilibrium is crossed at most 1 time. The motion can look like either sketch below.

Two sketches of displacement $x$ against time $t$ (the $x$-axis arrow points down, since down is positive). Left: $x$ starts away from equilibrium and decays toward $0$ without ever crossing it. Right: $x$ starts away from equilibrium, crosses equilibrium once, overshoots a little, and then decays back toward $0$ from the other side.

Case III: ${\lambda }^2-{\omega }^2<0$ (complex roots)

$$x(t)=e^{-\lambda t}[c_1\cos \left(\sqrt{{\omega }^2-{\lambda }^2}t\right)+c_2\sin \left(\sqrt{{\omega }^2-{\lambda }^2}t\right)]$$

This is underdamped. The bracket oscillates with amplitude $\sqrt{c_1^2+c_2^2}$, and the factor $e^{-\lambda t}$ shrinks it, so the motion stays inside the envelope $\pm \sqrt{c_1^2+c_2^2}{e}^{-\lambda t}$.

Displacement $x$ against time $t$ (the $x$-axis arrow points down). The curve oscillates back and forth across equilibrium, and its swings shrink over time. It stays between the two envelope curves $\pm \sqrt{c_1^2+c_2^2}{e}^{-\lambda t}$ (drawn in blue), which decay to $0$.

Driven motion

$${\displaystyle m\frac{d^{2}x}{d{t}^2}+kx+\beta \frac{dx}{dt}=f(t)}$$

The forcing term $f(t)$ makes the equation nonhomogeneous, so we use undetermined coefficients or variation of parameters.

5.1d

LRC-series circuit

${\displaystyle L\frac{di}{dt}+Ri+\frac{1}{C}q=E(t)}$, which is

$${\displaystyle L\frac{d^{2}q}{d{t}^2}+R\frac{dq}{dt}+\frac{1}{C}q=E(t)}$$

The first two terms, ${\displaystyle L\frac{di}{dt}+Ri}$, alone give an LR-series circuit. The last two, ${\displaystyle R\frac{dq}{dt}+\frac{1}{C}q}$, alone give an RC-series circuit.

“Underdamped,” “critically damped,” and “overdamped” have the same meanings as with spring/mass systems.

The complementary function is the transient term/solution, and the particular solution is the steady-state term/solution.

Example

Find the charge on the capacitor in an LRC-series circuit when ${\displaystyle L=\frac{1}{4}}$ h, $R=20\Omega$, ${\displaystyle C=\frac{1}{300}}$ f, $E(t)=0$ V, $q(0)=4$ C, and $i(0)=0$ A. Is the charge on the capacitor ever equal to zero?

Solution

Substitute the values (${\displaystyle \frac{1}{C}=300}$):

$${\displaystyle \frac{1}{4}\frac{d^{2}q}{d{t}^2}+20\frac{dq}{dt}+300q=0}$$

Multiply by 4 and factor the auxiliary equation:

$$q″+80q\prime +1200q=0⟹(m+60)(m+20)=0$$

$$m=-20,-60$$

$$\begin{array}{cc}\hfill {\displaystyle q(t)}& {\displaystyle =c_1{e}^{-20t}+c_2{e}^{-60t}}\hfill \\ \hfill {\displaystyle q\prime (t)}& {\displaystyle =-20c_1{e}^{-20t}-60c_2{e}^{-60t}}\hfill \end{array}$$

Since $i=q\prime$, the initial conditions give

$$\begin{array}{cc}\hfill {\displaystyle q(0)=4}& {\displaystyle ⟹c_1+c_2=4}\hfill \\ \hfill {\displaystyle q\prime (0)=0}& {\displaystyle ⟹-20c_1-60c_2=0}\hfill \end{array}$$

Divide the second equation by 20 to get $-c_1-3c_2=0$, then add it to $c_1+c_2=4$:

$$-2c_2=4⟹c_2=-2$$

Then $c_1=6$.

Does $q(t)=0$? Set $6e^{-20t}-2e^{-60t}=0$ and multiply by ${\displaystyle \frac{1}{2}{e}^{60t}}$:

$${\displaystyle 3e^{40t}-1=0⟹e^{40t}=\frac{1}{3}}$$

$${\displaystyle t=\frac{1}{40}\ln \frac{1}{3}<0}$$

The only zero is at a negative time.