- #1

- 12

- 0

I have a serious problem solving the following system of differential equations:

[tex]

m_1\ddot{x_1}=-k_1x_1-k(x_1-x_2)

[/tex]

[tex]

m_2\ddot{x_2}=-k_2x_2-k(x_2-x_1)

[/tex]

Does anybody have an idea how to solve it?

Thanks.

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- Thread starter ILens
- Start date

- #1

- 12

- 0

I have a serious problem solving the following system of differential equations:

[tex]

m_1\ddot{x_1}=-k_1x_1-k(x_1-x_2)

[/tex]

[tex]

m_2\ddot{x_2}=-k_2x_2-k(x_2-x_1)

[/tex]

Does anybody have an idea how to solve it?

Thanks.

- #2

- 13,172

- 741

It can be easily decoupled obtaing a 4-th order ODE in one of the coordinates.Which can be solved.

- #3

- 12

- 0

- #4

- 13,172

- 741

- #5

- 12

- 0

Thanks for your answer.

You have just confirmed that the solution I have derived is correct :rofl:

You have just confirmed that the solution I have derived is correct :rofl:

- #6

- 2,538

- 681

There is no need to get into a 4th order ODE.

- #7

Tide

Science Advisor

Homework Helper

- 3,089

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Laplace Transforms, anyone? :-)

- #8

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Let [tex] x_{i} = a_{i}e^{i\omega t} [/tex], substitute and get the following equations

[tex] -a_{1}m_{1}\omega^{2} = -a_{1}k_{1} - k(a_{1} - a_{2}) [/tex]

[tex] -a_{2}m_{2}\omega^{2} = -a_{2}k_{2} - k(a_{2} - a_{1}) [/tex]

Solve this set of equations for [tex] \omega [/tex] the use the boundary and initial conditions to obtain the [tex] a_{1} [/tex] and [tex] a_{2} [/tex]. This is a complicated set of equations because of the 3 distinct spring constants. at first glance it was exactly solvable almost trivial, but 3 constants makes it an order of magnitude more difficult.

[tex] -a_{1}m_{1}\omega^{2} = -a_{1}k_{1} - k(a_{1} - a_{2}) [/tex]

[tex] -a_{2}m_{2}\omega^{2} = -a_{2}k_{2} - k(a_{2} - a_{1}) [/tex]

Solve this set of equations for [tex] \omega [/tex] the use the boundary and initial conditions to obtain the [tex] a_{1} [/tex] and [tex] a_{2} [/tex]. This is a complicated set of equations because of the 3 distinct spring constants. at first glance it was exactly solvable almost trivial, but 3 constants makes it an order of magnitude more difficult.

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