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Damped SHM w/ forced motion

  • Thread starter xago
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  • #1
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Homework Statement



[PLAIN]http://img848.imageshack.us/img848/1150/damped.png [Broken]

Homework Equations





The Attempt at a Solution


Part i)

x(t) = C1coswt + C2sinwt
dx/dt = -C1wsinwt +C2wcoswt

using IC's:
xo = C1
dxo/dt = C2w

therefore x(t) = xocoswt + (dxo/dt)/w sinwt

but for part ii I do the same thing and get A=0 and B=0. I know I'm doing something wrong here what is it?
 
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Answers and Replies

  • #2
vela
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First off, the notation in the problem seems a bit sloppy to me. The two functions xs(t) and xp(t) refer to the same function. Also, the form of the homogeneous solution isn't correct if you take A and B to simply be constants.

Anyway, what you did is not how you solve for C1 and C2. You want to plug the particular solution into the LHS of the differential equation and then collect terms. You'll get (something)cos ωt + (something else)sin ωt, where the two coefficients will depend on C1, C2, and the various parameters in the equation. When you compare that to the RHS, you can see that you must have (something)=f0 and (something else)=0. You can solve these equations for C1 and C2.

(If you've already taken differential equations, you're just solving for the particular solution using the method of undetermined coefficients.)

Once you have that, we can go on to tackle finding the homogeneous solution xh(t).
 
  • #3
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Ok, thanks for your help, I followed through with it and got this, which seems kind of insane but I checked it a lot and it seems right.

[PLAIN]http://img859.imageshack.us/img859/1029/scan0008k.jpg [Broken]
 
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  • #4
vela
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That can't be correct. Both coefficients should be proportional to f0, and the units don't appear to work out either.

I just noticed there's a typo in the differential equation. The last term on the LHS is missing an x(t). It should be

[tex]\ddot{x}(t) + 2\zeta\omega_n\dot{x}(t)+\omega_n^2 x(t)=f_0\cos\omega t[/tex]

The coefficients will be

[tex]\begin{align*}
C_1 &= -\frac{f_0(\omega^2-\omega_n^2)}{(\omega^2-\omega_n^2)^2+(2 \omega\omega_n\zeta)^2} \\
C_2 &= \frac{f_0 (2 \omega\omega_n\zeta)}{(\omega^2-\omega_n^2)^2+(2 \omega\omega_n\zeta)^2}
\end{align*}[/tex]
 
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  • #5
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Wow, nice find, I just double checked and you're right about the missing x term. Interesting since the question was straight out of the textbook and the teacher hasn't mentioned anything about it when he assigned the questions. I've emailed him to post a note on it so hopefully other student's don't solve it the same way too.
 
  • #6
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Just redid the whole problem and got the same coefficients, thanks for all your help!
 
  • #7
vela
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Were you able to work out the rest of the problem?
 

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