Undamped forced oscillations

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Homework Statement


This is an example of an Undamped Forced Oscillation where the phenomenon of Pure Resonance Occurs.

Find the solution of the initial value problem:
x'' + 4 x = 8 sin(2 t) , x(0)=x'(0)=0



Homework Equations





The Attempt at a Solution



in class we were given the equation:
x''+(k/m)x=(F0/m)cos(ѡt)
and,
x_p= [(F0/m)/(ѡ0^2-ѡ^2)]cos(ѡt)
where ѡ0=sqrt(k/m)

However, in this equation ѡ=2 and ѡ0=sqrt(4)=2
so, the equation to find x_p fails... since ѡ0^2-ѡ^2 =0

I feel like I am just missing something here... anyone want to help me out?
 

Answers and Replies

  • #2
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This is covered in basic differential equation courses.

1) First find the solution to the homogeneous equation, this is called the complementary solution.

Homogeneous eq:
[tex]x'' + 4x = 0[/tex].

2) Assume that the solution is of the form [tex]x = e^{rt}[/tex], where r is some constant. Then [tex]x'' = r^{2}e^{rt}[/tex].

3) Substitute:
[tex]r^{2}e^{rt} + 4e^{rt} = 0[/tex]
This simplifies to [tex]r^{2} + 4 = 0[/tex] because [tex]e^{rt} > 0[/tex]. Solving this yields r = -2i. This means that the general solution is [tex]x = Asin2t + Bcos2t[/tex], A and B are unknown constants. Using the boundary condition we find that A = 0, B = 0.

4) Now find the solution to your original question by assuming that x is of the form [tex]x = Atsin(2t) + Btcos(2t)[/tex]. This is the particular solution.

5) Differentiate: [tex]x' = Asin(2t) + 2Atcos(2t) + Bcos(2t) - 2Btsin(2t)[/tex]
[tex]x'' = 2Acos(2t) + 2Acos(2t) - 4Atsin(2t) - 2Bsin(2t) - 2Bsin(2t) - 4Btcos(2t)[/tex]
[tex]= 4Acos(2t) - 4Atsin(2t) - 4Bsin(2t) - 4Btcos(2t)[/tex].

6) Plug in: [tex]4Acos(2t) - 4Atsin(2t) - 4Bsin(2t) - 4Btcos(2t) + 4Axsin(2t) + 4Btcos(2t)[/tex]
[tex]= 8sin(2t), 4Acos(2t) - 4Bsin(2t) = 8sin(2t), A = 2, B = 0[/tex]

7) The general solution is the particular solution and complementary solution:
[tex]x = 2tsin(2t)[/tex] (complementary solution was 0)
 
  • #3
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I think you made an error when finding A and B:

4Acos(2t)-4Bsin(2t) = 8sin(2t)
==> A=0 and B=-2?

So, x_p = -2tcos(2t)?
 
  • #4
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Yeah I did make that mistake. You can always check your answer by differentiating x and plugging it into the differential equation.
 
  • #5
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x(t)=-2tcos(2t) is not the correct answer. I checked the problem again and the initial conditions, and they are right.

I remember in class that to find the A and B of the complimentary solution, you wait to apply the initial conditions until after you find the particular solution as well. Like so:

x(t) = -2tcos(2t) + Acos(2t) + Bsin(2t)
x'(t)= 4tsin(2t) - 2cos(2t) - 2Asin(2t) + 2Bcos(2t)

So,
x(0)= 0+A+0=0 ==> A=0
x'(0)= 0-2-2(A)+2B=0
2B=2 ==>B=1

So,
x(t) = -2tcos(2t)+sin(2t)

This answer was correct. Can anyone explain why you have to wait to apply the initial conditions?
 
  • #6
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If you don't wait you make an error like I did. When I applied the init conditions after solving for the complementary solution it caused the complementary solution to become zero.
 

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