EGR 1101: Unit 13 Lecture #1 Second-Order Differential Equations in Mechanical Systems (Section 10.5 of Rattan/Klingbeil text)
Review: Procedure Steps in solving a linear ordinary differential equation with constant coefficients: 1. Find the transient solution. 2. Find the steady-state solution. 3. Find the total solution by adding the results of Steps 1 and Apply initial conditions (if given) to evaluate unknown constants that arose in the previous steps.
Forcing Function = 0? Recall that if the forcing function (the right- hand side of your differential equation) is equal to 0, then the steady-state solution is also 0. In such cases, you get to skip straight from Step 1 to Step 3!
Second-Order Equations
Imaginary Numbers? In solving some second-order linear ordinary differential equations with constant coefficients, you’ll get imaginary numbers as you work through Step #1 (transient solution). To simplify your solution in such situations, use Euler’s identity:
Today’s Examples 1. Free vibration of a spring-mass system 2. Forced vibration of a spring-mass system
MATLAB Commands for Example #2 >>fplot('1/(1-0.9^2)*(cos(0.9*t)-cos(t))', [0 200]) To investigate the system’s behavior as the forcing frequency approaches the system’s natural frequency, change the two occurrences of 0.9 in this command to 0.99, and then change to
Resonance at work Glass shattered by resonance: Tacoma Narrows Bridge collapse:
EGR 1101: Unit 13 Lecture #2 Second-Order Differential Equations in Electrical Systems (Section 10.5 of Rattan/Klingbeil text)
Low-Pass and High-Pass Filters A low-pass filter is a circuit that passes low-frequency signals and blocks high- frequency signals. A high-pass filter is a circuit that does just the opposite: it blocks low-frequency signals and passes high-frequency signals.
Today’s Examples 1. Second-order low-pass filter
Mechanical-Electrical Analogy Governing equation of forced spring-mass system from previous lecture: Initial conditions: Governing equation of our second-order filter: Initial conditions:
Mechanical-Electrical Analogy (Continued) Solution of forced spring-mass system from previous lecture: Solution of our second-order filter: