Real Eigenvalues: Answer to Example1

Example: Consider the harmonic oscillator with spring constant , damping constant , and the mass m=1.

(1)

Write down the second order equation governing this physical system. Use the letter y for the spring's displacement from its rest position.

(2)

Convert this equation into a linear system of first order differential equations.

(3)

Solve the system.

(4)

Find the particular solution which satisfies the initial conditions

(5)

Discuss the long-term behavior of the system. Was this conclusion probable?

Answer:

(1)

The differential equation is

.

Using the values for the constants, we get

.

(2)

Set y'=v, then we have

.

Hence, we have the system

;

(3)

In order to solve the above system, we first need to find the eigenvalues of the system. Note that the matrix coefficient is

.

The characteristic equation is given by

.

Its roots are

,

which gives

For every eigenvalue, we need to find an eigenvector.

• . Let V be an associated eigenvector such that

  .

  The vector V must satisfy the system of algebraic equations

  

  Clearly, the two equations reduce to the same equation

  .

  Hence, we have

  .

  We choose

  .

• . Let V be an associated eigenvector such that,

  .

  The vector V must satisfy the system of algebraic equations

  

  Clearly, the two equations reduce to the same equation

  .

  Hence, we have

  .

  We choose

  .

Now we are ready to write down the general solution

,

where

.

(4)

In order to find the solution to the harmonic oscillator system which satisfies the initial conditions y(0)= 0 and y'(0)=1, we need the general solution which gives y. From the general solution to the system we get

,

and . The equation giving v is obvious and can be obtained from y since v=y' (you may want to check that we did not make any mistakes). The initial conditions imply

Solving it we get

.

Therefore, the solution is

(5)

The long-term behavior of the solution is now obvious since

,

meaning that the system tends to its rest position. Note that since the eigenvalues are both negative, it was clear from the outset that the solution will tend to its unique equilibrium position.

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