Base excitation of undamped Spring-mass system

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SUMMARY

The discussion centers on solving a base excitation problem in an undamped spring-mass system using differential equations (DE) and Laplace Transforms. The user, Rian, successfully formulated the DE and derived the solution for y(t) using the Laplace Transform, but seeks clarification on the characteristic equation and frequency response. Key variables include F0 (amplitude of base excitation), ω (forcing frequency), and ω0 (natural frequency). The conversation emphasizes the importance of understanding the linearity of the DE and the distinction between transient and steady-state responses.

PREREQUISITES
  • Understanding of differential equations, particularly second-order non-homogeneous equations.
  • Familiarity with Laplace Transforms and their application in solving DEs.
  • Knowledge of frequency response analysis in dynamic systems.
  • Concept of characteristic equations in the context of linear systems.
NEXT STEPS
  • Study the derivation and application of characteristic equations for second-order differential equations.
  • Learn about the Fourier Transform and its role in analyzing frequency response.
  • Explore the concept of transient versus steady-state responses in undamped and damped systems.
  • Investigate the implications of resonance in spring-mass systems and methods to mitigate it.
USEFUL FOR

Students and professionals in engineering, particularly those focused on mechanical systems, control theory, and signal processing, will benefit from this discussion.

Rian199
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Homework Statement


I am a little stuck/unsure in an homework question for my signals and systems course. See question sheet attached. I have been able to do question 1 for the first section using FBD, and I have solved the DE using Laplace Transforms (See below for my solution). I am both unsure if my solution is correct, and what to do for question 2 and 3 of Section 1A. It might be because I did not learn most of this is English, but I am just not sure exactly what those two questions are asking?Please advise, thank you! Just a place to start at the very least would suffice!

Homework Equations

The Attempt at a Solution


y(t) = [(F0 * ω02)]/[(ω02 - ω2)] * sin(ωt) + C1*cos(ω0t) + C2*sin(ω0t)
 

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Hello again, Rian,

Well, at least the problem statement is legible now :smile:
Part 1 asks to show something, but you ask if your solution of the DE is correct (without showing how you got it, so you make it a little difficult). What is ##F_0, \ \omega, \ \omega_0 \ \ ## ? (I know, but you apparently assume some things you don't mention)

For part 2: you know (or can google) what a characteristic equation for a differential equation is. Can you confirm ?

For part 3: the frequency response is (amplitude of y) divided by (amplitude of h) as a function of ##\omega##. Clearly things become problematic at some value of ##\omega##. Something to fix that is attempted on page 2, so you have a long way to go after Section 1A.

The subject at hand is very interesting and extremely useful in many fields of science !
 
BvU said:
Hello again, Rian,

Well, at least the problem statement is legible now :smile:
Part 1 asks to show something, but you ask if your solution of the DE is correct (without showing how you got it, so you make it a little difficult). What is ##F_0, \ \omega, \ \omega_0 \ \ ## ? (I know, but you apparently assume some things you don't mention)

For part 2: you know (or can google) what a characteristic equation for a differential equation is. Can you confirm ?

For part 3: the frequency response is (amplitude of y) divided by (amplitude of h) as a function of ##\omega##. Clearly things become problematic at some value of ##\omega##. Something to fix that is attempted on page 2, so you have a long way to go after Section 1A.

The subject at hand is very interesting and extremely useful in many fields of science !

Thanks for the advice regarding posting. Well for the DE, I just proved it using a FBD, and I'm confident the actual DE is solid. I then asumed:

h(t) = F0cos(ωt)

ω ≡ Forcing frequency
ω0 ≡ Natural Frequency of the system
F0 ≡ Amplitude of the base excitation

I took the laplace transform of of each term, rearranged for Y(s) and then did the inverse transform to get y(t).

For part 2, correct me if I'm wrong, but I know normally the characteristic equation of a DE is when you arrange the DE in standard form and then use the coefficients ans polynomials to find the roots?
I've just been unable to figure out how to do it with a 2nd order non-homogeneous function like this one?

So with part 3, should I then be taking the Fourier transform of each term for my solution for y(t) and h(t), treating h(t) as an input and dividing the two? If so, how do I find the amplitude of each? Yes, I get that much, at resonance the transient part goes to infinity (ω=ω0).
I know is still have a bit to go, but I feel like if I can grasp this bit, it sort off repeats itself in the next parts? Thank you again.
 
Also, for the characteristic equation, even though it's non-homogeneous, is that just referring to the characteristic equation of the homogeneous variant when your finding the solution to the complimentary solution (DC)?
 
Also, in that case the homog soln is:

assuming y(t) = eλt:
y''(t) = λ2eλt

thus

λ2eλt + ω02eλt = 0

λ = ±iω0?
 
Part 1 is just fine. Part two too: yes, for the characteristic equation you look at the homogeneous equation (and indeed find ##\omega_0## as you did)

For part 3 you make use of the linearity of the equation: if f(t) is a solution of the homogeneous equation and g(t) one particular solution to the inhomogeneous equation, then f(t) + g(t) is a general solution to the equation + initial conditions. The initial conditions (your C1 and C2) can be applied to f(t) .

The g(t) that goes to infinity in the undamped case is not the transient part. That term will become clearer when you consider the damped case: it is the f(t) that is the transient response; the C1 and C2 are irrelevant for the long term. But in the undamped case the transient response isn't "transient" :smile:
 
Gotchya! Yea sorry I misspoke, and I realized after posting that the undamped condition is unbounded and will go to infinity without the dampener in the improved case. I'll have a further crack at it and post back afterwards.
 
BvU said:
Part 1 is just fine. Part two too: yes, for the characteristic equation you look at the homogeneous equation (and indeed find ##\omega_0## as you did)

For part 3 you make use of the linearity of the equation: if f(t) is a solution of the homogeneous equation and g(t) one particular solution to the inhomogeneous equation, then f(t) + g(t) is a general solution to the equation + initial conditions. The initial conditions (your C1 and C2) can be applied to f(t) .

The g(t) that goes to infinity in the undamped case is not the transient part. That term will become clearer when you consider the damped case: it is the f(t) that is the transient response; the C1 and C2 are irrelevant for the long term. But in the undamped case the transient response isn't "transient" :smile:
So this is what I got. Did both part 1 and 2 (See attached file), I am just unsure regarding Part 3 still?
 

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