Sunny129
Diamond Member
- Nov 14, 2000
- 4,823
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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 abviously 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 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 abviously 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 FO/m cos(wt + B)
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