Question

Question on Aging Spring with Damping ?_?

Answer 1

These words need decoding.

Answer 2

As a spring ages, it's spring "constant" decreases in value. One such model for a mass spring system with an aging spring is:

Answer 3

\[mx''(t)+bx'(t)+ke^{-\eta t}x(t)=0\]

Answer 4

where m is the mass, b is the damping constant, k and n positive constants, and x(t) the displacement of the spring from it equilibrium position. Let m=1 kg, b=2 N-sec/m, k=1 N/m, and n=1(sec)^-1. The system is set in motion by displacing the mass 1, from its equilibrium position and then releasing it (x(0)=1,x'(0)=0). Find at least the first four nonzero terms in a power series expansion about t=0 for the displacement.

Answer 5

I can't remember all of it but you need to find its power series expansion, differentiate it, then look at terms and get four no-zero coefficiens of these terms.

Answer 6

I just need help setting up the equation.

Answer 7

second order ODE :

\(x'' + 2x' + 2^{-t}x = 0\)
\(x(0) = 1, x'(0) = 0\)

Answer 8

thank you but does this look right to you?

Answer 9

wait a second! there's no E in the problem whoops, but also why did you put 2 front of ^-t, shouldn't it be k=1?? I'm confused

Answer 10

ahh sorry it was by mistake
\(x'' + 2x' + e^{-t}x = 0\) is the correct eq'n

Answer 11

wait oh my god no! is there really an e??

Answer 12

oh shoot there is...

Answer 13

yes lol

Answer 14

\(e^{-x} = 1 - x + \dfrac{x^2}{2!} - \dfrac{x^3}{3!} + \dfrac{x^4}{4!} - ... \)

Answer 15

i think u may use that maclaurin expansion about 0

Answer 16

i'm so silly nevermind thank you... ;_; (crying face)

Answer 17

lol np :) btw i never worked on series solutions for diff.eq'ns before... i might be a bit off lol... care to explain quick how these work ?

Answer 18

http://tutorial.math.lamar.edu/Classes/DE/SeriesSolutions.aspx

Answer 19

well for this problem we only want the first four nonzero terms in the series, so we just plug in the first few n=1,2,3 and solve for a_1,a_2,a_3 like in my picture.

Answer 20

ohk im going thru that link... ty :)