Hard Limit Problem: Damped Harmonic Oscillator

[Eric_meyers]

Homework Statement

"Demonstrate that in the limit (nu)---> 2w0 the solution to the damped harmonic oscillator equation becomes x(t) = (x0 + [v0 +(nu/2)x0]t)e^(-(nu)t/2)

Homework Equations

Solution to damped harmonic oscillator equation; x(t) = a*e^(-(nu)t/2)cos(w1t - (theta))

Where a = [x0^2 + (v0 + (nu)x0/2)^2/w1^2]1/2 and

(theta) = tan^-1((v0 + (nu)x0/2)/w1x0)

and w1^2 = w0^2 - (nu)^2/4

The Attempt at a Solution

So I plugged (nu) ---> 2w0 into my equations and got ...

x0^2 + [(v0 + w0x0)^2)/(w0^2 - w0^2)]^1/2 * e^(-w0t) * cos (-tan^-1 (( v0 + w0x0)/((w0^2-w0^2)^1/2*x0))

and I note that the divide by 0 in my tan inverse goes to infinity which equals pi/2 and the cos of pi/2 is 0

but then before my e term I have (w0^2 - w0^2) on the denominator which means that's going to infinity... so I have one part going to 0 and one part going to infinity...

and I don't know how to deal with this.

 

[anathema]

I would approach this problem by first verifying that the given solution is indeed a valid solution to the damped harmonic oscillator equation. I would also check if the limit (nu) ---> 2w0 is physically meaningful and if it is within the range of applicability for the equation.

Assuming that the given solution is correct and the limit is valid, I would then analyze the behavior of each term in the solution as (nu) approaches 2w0. For example, as you mentioned, the term (v0 + (nu)x0/2) in the numerator of the expression for a goes to infinity, while the denominator (w1) also goes to infinity. This suggests that the limit of a is indeterminate and requires further investigation.

Next, I would consider the behavior of the cosine term in the solution. As (nu) approaches 2w0, the argument of the cosine function becomes (-tan^-1 (infinity)), which can be rewritten as (-pi/2). This means that the cosine term approaches 0, which is consistent with the solution given in the forum post.

Finally, I would consider the overall behavior of the solution as (nu) approaches 2w0. Since the cosine term goes to 0 and the exponential term goes to infinity, the overall behavior of the solution depends on the behavior of the term in the parentheses [x0 + (v0 + (nu)x0/2)t]. This term approaches x0 as (nu) approaches 2w0, which means that the solution approaches x(t) = x0e^(-(nu)t/2), which is similar to the given solution but without the additional term in the parentheses.

Overall, the behavior of the solution as (nu) approaches 2w0 suggests that the solution may not be valid for this specific limit. It is possible that the equation is only applicable for a certain range of (nu) values and the given limit falls outside of that range. I would recommend further investigation and possibly consulting with other experts in the field to determine the validity of this solution for the given limit.

 

[tomcue]

As (nu) approaches 2w0, the damped harmonic oscillator equation becomes more and more heavily damped, resulting in a decrease in amplitude and a shift in phase angle. This can be seen in the solution provided, where the amplitude is directly affected by (nu) and the phase angle is affected by both (nu) and w0. As (nu) approaches 2w0, the amplitude decreases to 0 and the phase angle approaches pi/2, resulting in a solution of x(t) = 0. This is expected in heavily damped systems, where the oscillations eventually die out completely. The fact that the denominator in the phase angle goes to infinity is not an issue, as it is multiplied by a term that goes to 0, resulting in a well-defined limit. Therefore, the solution provided is a valid representation of the damped harmonic oscillator equation in the limit as (nu) approaches 2w0.