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eku_girl83
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Driven, damped harmonic oscillator -- need help with particular solution
Consider a damped oscillator with Beta = w/4 driven by
F=A1cos(wt)+A2cos(3wt). Find x(t).
I know that x(t) is the solution to the system with the above drive force.
I know that if an external driving force applied to the oscillator then the total force is described by F = -kx - bx' + F0cos(wt).
But in our case the driving force is A1cos(wt)+A2cos(3wt) so
F=-kx-bx+A1cos(wt)+A2cos(3wt).
Then our differential equation is mx''+bx'+kx=A1cos(wt)+A2cos(3wt).
This can also be written as x''+2Betax'+(w^2)x=A1cos(wt)+A2cos(3wt).
For the complementary solution, we set the right side of the equation equal to zero and solve for x. This is o.k.
However, I am having trouble with the particular solution. Can someone tell me how I find a particular solution for this? I can find the particular solution for x''+2Beta x'+ (w^2)x = A cos (wt), but what about the particular solution when the driving force is not A cos (wt), as we have in this case?
Any help GREATLY appreciated!
Consider a damped oscillator with Beta = w/4 driven by
F=A1cos(wt)+A2cos(3wt). Find x(t).
I know that x(t) is the solution to the system with the above drive force.
I know that if an external driving force applied to the oscillator then the total force is described by F = -kx - bx' + F0cos(wt).
But in our case the driving force is A1cos(wt)+A2cos(3wt) so
F=-kx-bx+A1cos(wt)+A2cos(3wt).
Then our differential equation is mx''+bx'+kx=A1cos(wt)+A2cos(3wt).
This can also be written as x''+2Betax'+(w^2)x=A1cos(wt)+A2cos(3wt).
For the complementary solution, we set the right side of the equation equal to zero and solve for x. This is o.k.
However, I am having trouble with the particular solution. Can someone tell me how I find a particular solution for this? I can find the particular solution for x''+2Beta x'+ (w^2)x = A cos (wt), but what about the particular solution when the driving force is not A cos (wt), as we have in this case?
Any help GREATLY appreciated!