Professor Kate Poirier | OL67 | Fall 2020

Author: Brian Enriquez (Page 1 of 5)

Project #4: Tacoma Narrows Bridge Collapse and Differential Equations

Forgive the smooth brain energy in this post. I’m doing this with less than a day left before the deadline. Anyway…


The Tacoma Narrows Bridge is a suspension bridge in Washington state, and was first constructed in 1937. It became infamous for collapsing in 1940 due to issues with vibrations (which can be modeled or analyzed with differential equations). This collapse prompted further studies into structural design and its relationship with vibrations.

Best Tacoma Bridge Collapse GIFs | Gfycat
Tacoma Narrows bridge oscillating in response to wind before collapsing

Bridges are designed to be somewhat flexible to accommodate the expansion and contraction of materials due to temperature changes and to survive vibrations caused by wind, moving loads (moving pedestrians/cars), and earthquakes . Like tuning forks, structures have natural frequencies (naturally vibrating at a specific rates when disturbed). At the day of the Tacoma Narrows Bridge collapse, the bridge was reacting to strong winds that eventually matched the natural frequency of the bridge, causing greater oscillations and stress on to the bridge. The bridge was described to be twisting at each end in opposite directions.

There’s more physics involved in this, but I’m not really capable of explaining in detail. Winds make bridges wobble. Too much wobble equals broken bridge. Stiff bridge with too much wobble equals broken bridge too. Too flexible bridge makes for bad driving experience.

Relevance to Differential Equations

When studying second order linear differential equations, we define f(x) in y''+p(x)y'+q(x)y=f(x) as the “forcing function.”

In practical applications, the “forcing function” is the force affecting a system. When the forcing function matches a natural frequency, “resonance” is achieved and we find that the amplitude of the oscillations increase. So for a suspension bridge, under certain conditions, the wind will “excite” the structure due to resonance, inducing strenuous movement that will lead to structural failure.

To counteract (excessive) oscillations due to external disturbances, structures have systems that “damp” vibrations. Suspension bridges can be seen as a series of spring-mass systems, therefore, the concept of “damping” in spring-mass systems can be applied to suspension bridges. Remember that the roots of the characteristic equation of the differential equation describing a spring-mass system determine the level of “damping” in a system (underdamped, overdamped, critically damped).

The following equation is a model developed by a pair of mathematicians (Lazer and Kenna) to describe the simple behavior of a suspension bridge with no wind disturbances:

\frac{d^2y}{dt^2}+\alpha\frac{dy}{dt}+\beta y+c(y)=-g

Notice that this equation is a second order differential equation (which we might be able to consider as a linear equation, if the forcing function is f(x)=-g-c(y)

\frac{d^2y}{dt^2} is the vertical acceleration of the bridge’s roadbed. \alpha\frac{dy}{dt} is derived from damping.\beta y “accounts for the force provided by the material of the bridge” pulling the roadbed to it’s neutral position. c(y) “accounts for the pull of the cable” when the roadbed is displaced in either upward or downward direction.-g is the downward force due to gravity. \alpha and \beta are constants.

For conditions where wind disturbance is a factor, the model becomes:

\frac{d^2y}{dt^2}+\alpha\frac{dy}{dt}+\beta y+c(y)=-g+\lambda \sin(\mu t)

\lambda \sin(\mu t) models a simplified oscillation caused by the wind.



No MLA or APA format but here are some links

Why the Tacoma Narrows Bridge Collapsed: An Engineering Analysis

Sound Waves and Music – Lesson 4 – Resonance and Standing Waves: Natural Frequency

5.4 Forced vibration of damped, single degree of freedom, linear spring mass systems.

Differential Equations:4th Edition by Paul Blanchard/Robert L. Devaney/Glen R. Hall

Help me. My brain is fried.

Test #3 Review: WebWork: Practice Circuits: Problem 3


A series circuit has a capacitor of {\textstyle\frac{1}{48}} \times 10^{-6} F and an inductor of 3 H. If the initial charge on the capacitor is 2 \times 10^{-2} C and there is no initial current, find the charge on the capacitor Q(t).


Capacitance, C = {\textstyle\frac{1}{48}} \times 10^{-6}

Inductance, L = 3 H

Initial Charge, Q(0) = Q_0 = 2 \times 10^{-2} C

Initial Current, Q'(0) = Q'_0 or I(0) = I_0 = 0 A

Since a resistor was not mentioned to be a part of the circuit, we can assume that R = 0  \Omega.

Also, there’s no impressed voltage mentioned, therefore, we can assume that E(t) = 0.

Using the form

LQ''+RQ'+\frac{1}{C}Q = E(t)

The differential equation for this problem will be

3Q''+(\frac{1}{({\textstyle\frac{1}{48}} \times 10^{-6})})Q = 0

The characteristic equation will then be

3r^2 + (\frac{1}{({\textstyle\frac{1}{48}} \times 10^{-6})}) = 0

Using the quadratic formula, the roots of the characteristic equation is

r = 0 \pm 4000i

Since the roots are complex imaginary numbers, the general equation for the amount of charge in the circuit will then be

Q(t) = e^{0t}(c_{1}sin(4000t)+c_{2}cos(4000t))

= c_{1}sin(4000t)+c_{2}cos(4000t)

Also note that, since the roots are imaginary, the oscillation of the circuit is underdamped.

To find the function for current, differentiate the previous equation, yielding

I(t) = Q'(t) = 4000(c_{1}cos(4000t)-c_{2}sin(4000t))

Next, substitute the initial conditions to solve for the undetermined coefficients

c_{1}sin(4000(0))+c_{2}cos(4000(0)) = 2 \times 10^{-2}

c_{2} = 2 \times 10^{-2}

4000(c_{1}cos(4000(0))-c_{2}sin(4000(0))) = 0

c_{1} = 0

Finally, the equation for the amount of charge in the given circuit is

Q(t) = 2 \times 10^{-2}cos(4000t)

Project#3: Section 6.3: Problem 6


(click the link to view video)



Use “Undetermined coefficients” method

  1. Solve for the roots of the characteristic equation.
    • use quadratic formula
    • use the the roots of the characteristic equation to find the solution of the complimentary equation
      • refer back to solutions of “constant coefficient second order linear equations”
  2. Find a suitable general form for a “particular solution”, y_{p} , that matches the “right side of the given equation” (y_{p} is the same as Q_{p}, which is function representing the amount of charge of a sytem).
    • if the right side of the equation is cos(ct), sin(ct), or cos(ct)+sin(ct), then the particular solution is of the form Acos(ct)+Bsin(ct)
    • since no terms in y_{p} is a solution of the complimentary equation, no extra steps are necessary in deciding the form of the particular solution
  3. Differentiate the chosen particular solution twice to find y_{p}^' and y_{p}^''.
  4. Substitute the values of y_{p} , y_{p}^' , and y_{p}^'' into Q, Q', and Q''.
  5. Simplify the new equation.
  6. Rearrange the terms of the equation (factoring may be necessary) so that the “left side of the equation” matches the form of the “right side of the equation”.
  7. Create and solve a system of equations to solve for the undetermined coefficients.
  8. Plug in the values of the undetermined coefficients into the chosen particular solution to determine Q_{p}.
  9. Since the steady steady state current, I_{p} , is equal to Q_{p}^' , we must differentiate Q_{p} to solve for the steady state current of the system.
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