How Does the Driven Oscillator ODE Describe Long-Term Motion?

[Eric_meyers]

Homework Statement

"The equation mx'' + kx = F0 * Sin (wt) governs the motion of an undamped harmonic oscillator driven by a sinusoidal force of angular frequency w. Show that the steady-state solution is

x = F0 * Sin (wt) /(m * (w0^2 - w^2))

Homework Equations

x(t) = xta(t) + xtr(t) where xta = long term behavior and xtr = transient piece of solution

xta(t) x0 cos (wt - (phi)

where x0 = w0^2 X0/[(w0^2 - w^2)^2 + v^2*w^2]^1/2

and phi = tan^-1(vw/(w0^2 - w^2)

The Attempt at a Solution

Honestly I'm not sure if the above equations are necessary. All my lecture notes tell me is that the long term behavior follows x0 * cos(wt - (phi)) so I'm not sure how I'm suppose to take that and turn it into F0 * Sin (wt) /(m * (w0^2 - w^2))

I mean obviously I have a sin force driving my harmonic oscillator, but how do I use that to determine my long term solution?

I really can't go much further than this at this point, I've been looking at this problem for a long time now and I can't scrap together anything useful.

 

[Redbelly98]

Homework Statement

"The equation mx'' + kx = F0 * Sin (wt) governs the motion of an undamped harmonic oscillator driven by a sinusoidal force of angular frequency w. Show that the steady-state solution is

x = F0 * Sin (wt) /(m * (w0^2 - w^2))

How about substituting this expression for x, and also the expression for x'', into the original differential equation?

 

[tomcue]

Dear Student,

The equation you provided is known as the driven oscillator ODE, and it governs the motion of an undamped harmonic oscillator driven by a sinusoidal force of angular frequency w. It is important to note that this equation is a second-order differential equation, which means that it has two solutions: a transient solution and a steady-state solution. The transient solution describes the initial behavior of the system, while the steady-state solution describes the long-term behavior.

To find the steady-state solution, we can use the principle of superposition, which states that the total solution of a linear system is the sum of its individual solutions. In this case, the total solution can be written as the sum of the transient solution and the long-term solution:

x(t) = xtr(t) + xta(t)

where xtr is the transient solution and xta is the long-term solution. We can rewrite the equation as:

x(t) = x0 cos (wt - (phi)) + xta(t)

where x0 is the amplitude of the transient solution and phi is the phase angle. Now, let's focus on the long-term solution, which is given by:

xta(t) = F0 * Sin (wt) /(m * (w0^2 - w^2))

where F0 is the amplitude of the driving force, m is the mass of the oscillator, w0 is the natural frequency of the oscillator, and w is the frequency of the driving force.

To derive this solution, we can use the method of undetermined coefficients. This method allows us to find a particular solution that satisfies the given equation. In this case, we can assume that the long-term solution has the form:

xta(t) = A * Sin (wt) + B * Cos (wt)

where A and B are constants to be determined. Substituting this into the original equation, we get:

mx'' + kx = F0 * Sin (wt)

which simplifies to:

mw^2(A * Sin (wt) + B * Cos (wt)) + k(A * Sin (wt) + B * Cos (wt)) = F0 * Sin (wt)

Equating the coefficients of Sin (wt) and Cos (wt), we get two equations:

mw^2A + kB = 0

and

-kA + mw^2B = F0

Solving these equations for A and B, we get:

A