# Undamped forced oscillations

## 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

## 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?

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:
$$x'' + 4x = 0$$.

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

3) Substitute:
$$r^{2}e^{rt} + 4e^{rt} = 0$$
This simplifies to $$r^{2} + 4 = 0$$ because $$e^{rt} > 0$$. Solving this yields r = -2i. This means that the general solution is $$x = Asin2t + Bcos2t$$, 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 $$x = Atsin(2t) + Btcos(2t)$$. This is the particular solution.

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

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

7) The general solution is the particular solution and complementary solution:
$$x = 2tsin(2t)$$ (complementary solution was 0)

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)?

Yeah I did make that mistake. You can always check your answer by differentiating x and plugging it into the differential equation.

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?

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.