Physics Help Part 2

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thEnEuRoMancER

Golden Member
Oct 30, 2000
1,415
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ad Problem 1, A

The general solution of your differential equation has the following form:

x(t)=D*exp(-zeta*omegan*t)*sin(omegad*t+phi)+E*sin(omega*t-psi)

(first part of the solution is the transient and second part is the steady state response),
where:

D - amplitude of the transient part
omegan - natural frequency, omegan = sqrt(m/n)
zeta - damping factor, zeta=b/(2*m*omegan)
omegad - damped natural frequency, omegad=sqrt(1-zeta^2)*omegan
phi - phase shift of the transient part
E - amplitude of steady state response, E = Fo/(sqrt((k-m*omega^2)^2+(b*omega)*2))
psi - phase shift of the steady state response, psi=arc tan (b*omega/(k-m*omega^2))

Equation of the function on the graph is:

x(t)=3*exp(-0,4394*t)*sin(4*pi*t-pi/2)+3*sin(((2*pi)/5)*t+pi()/2)

That's all I can do for you right now. Hope it helps.

 
Triumph

Triumph

Lifer
Oct 9, 1999
15,031
14
81
there are equations for answering all the questions for the first one. i don't remember what they are, but just remember ummm, i can't remember, and i ain't looking it up. but it's gotta be in your book. guess i'm not much help.
 
A

acejj26

Senior member
Dec 15, 1999
886
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Just Newtonian mechanics...the problem is that the book doesn't cover damped/driven SHO and our prof wants us to do this without his help - just rely on his notes, in which he makes at least 5 mistakes per class and makes taking notes almost pointless. I appreciate everyone's help.
 
Sunny129

Sunny129

Diamond Member
Nov 14, 2000
4,823
6
81
I wasnt sure if you got my post if your first thread, so i thought i would add it to this one too incase you arent checking your first thread anymore...


well the first problem is obviously a differential equation. to solve for the inhomogenious solution you must set the driving force = to zero. once yoou have your inhomogenious solutions, replace the driving force given in the original problem and solve. the final solution for x will be the sum of the inhomogenious and homogenious solutions...i dont know if that helps any. those are just the basics to solving a differential dealing with a driving force and simple harmonic motion. now if you can only turn these words into numbers and find a nice
solution...lord knows i can't...i hope this helped some...

EDIT, i don't know how to do part B (continuing the graph), and also, i'm taking a look at your work right now...no promises that i can tell if its right or not, but i'll try

2nd edit, i also mixed up the words homogenious and inhomogenious. the homogenious has no driving force, while the inhomogenious does.


x" + (b/m)x' + (k/m)x = FO/m cos(wt + B)

homogenious solution:

x" + (b/m)x' + (k/m)x = 0

r^2 + (b/m)r + (k/m) = 0

[-b +- (b^2 - 4ac)^(1/2)] / 2a

[-(b/m) +- ((b/m)^2 - 4(1)(k/m))^(1/2)] / 2(1)

[-(b/m) +- (b^2/m^2 -4k/m)^(1/2)] / 2

r = [-(b/m) +- ((b^2-4k/m) / m^2)^(1/2)]/ 2

this obviously would have been easier had your teacher given you a problem like x" - x' - 2x = 0. in this case you would have nice round numbers to use and the quadratic equation would not have to be used to solve for r:

r^2 - r - 2 = 0

(r + 1)(r - 2) = 0

r = -1, 2

so the solution to this homogenious (or undriven equation, where the driving force = zero) would be C1e^(-t) + C2e^(2t), where C1 and C2 are constants.

unfortunately, your teacher gave you the coefficients b/m and k/m, which dont foil nicely while using a quadratic equation, so i'm not sure you can write the solution to the
homogenious equation like so:

C1e^[[-(b/m) + ((b^2-4k/m) / m^2)^(1/2)]/ 2] and

C2e^[[-(b/m) - ((b^2-4k/m) / m^2)^(1/2)]/ 2]

and i hate to make things worse, but the problem only gets much more complicated when you solve for the inhomogenious solution(s) by replacing the right side of the equation with the following driving force:

FO/m cos(wt + B)
 
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