# Steady State Solution of Forced, Damped Harmonic Oscillator

• transmini
In summary: I missing something?You are correct in that cosine and sine are combined into a sine term. However, this equation is incorrect because the phase shift is incorrect. The correct equation is ##A(\omega)cos(\omega t+\phi)##.
transmini

## Homework Statement

A damped harmonic oscillator is driven by an external force of the form $$F_{ext}=F_0sin(\omega t)$$
Show that the steady state solution is given by $$x(t)=A(\omega)sin(\omega t-\phi)$$
where $$A(\omega)=\frac{F_0/m}{[(\omega_0^2-\omega^2)^2+4\gamma^2\omega^2]^{1/2}}$$
and $$tan\phi=\frac{2\gamma\omega}{\omega_0^2-\omega^2}$$

## Homework Equations

Equation of motion: $$\ddot{x}+2\gamma\dot{x}+\omega_0^2x=\frac{F_0}{m}sin(\omega t)$$

## The Attempt at a Solution

Using a characteristic equation ##\lambda^2+2\gamma\lambda+\omega_0^2## I found the solution to the equation of motion to be $$x(t)=Acos(\omega t)+Bsin(\omega t)$$ with another term that goes to zero at steady state, which we are considering only steady state. ##A## and ##B## are both arbitrary constants.

Then using the steady state equation found, I plugged this back into the equation of motion to find the coefficients.
$$\dot{x} = -A\omega sin(\omega t)+B\omega cos(\omega t)$$
$$\ddot{x} = -A\omega^2cos(\omega t)-B\omega^2sin(\omega t)$$
so
$$-A\omega^2cos(\omega t)-B\omega^2sin(\omega t)-2\gamma A\omega sin(\omega t)+2\gamma B\omega cos(\omega t)+\omega_0^2Acos(\omega t)+\omega_0^2Bsin(\omega t) = \frac{F_0}{m}sin(\omega t)$$
grouping sines and cosines:
$$(-A\omega^2+2\gamma B\omega+\omega_0^2)cos(\omega t)+(-B\omega^2-2\gamma A\omega+\omega_0^2B)sin(\omega t) = \frac{F_0}{m}sin(\omega t)$$

Since the coefficients for cosines on the left have to match the ones on the right, similarly for the sines, I got
$$-A\omega^2+2\gamma B\omega+\omega_0^2A = 0$$
$$-B\omega^2-2\gamma A\omega+\omega_0^2B=\frac{F_0}{m}$$

Then solving for the coefficients A and B, I got
$$A=\frac{-2F_0\gamma\omega}{m(\omega_0^2-\omega^2)^2+4m\gamma^2\omega^2}$$
$$B=\frac{F_0(\omega_0^2-\omega^2)}{m(\omega_0^2-\omega^2)^2+4m\gamma^2\omega^2}$$

Which when combining the sine and cosine term into ##A(\omega)cos(\omega t-\phi)## where ##A(\omega)=\sqrt{A^2+B^2}## and ##tan(\phi)=\frac{B}{A}##,

I receive exactly what I was supposed to get with the exception that I get $$tan(\phi)=\frac{-2\gamma\omega}{\omega_0^2-\omega^2}$$ rather than the positive version of the argument in the trig function.

I got told that "the method is wrong" and the " 'A' you received is not the same as the one we want". I don't see how the method is wrong, seeing as how it does in fact give a steady-state solution to the equation of motion (I did verify this to be sure). When using complex roots you get the correct signs on all the terms, but where did I go wrong with solving the differential equation using this method? If this method actually does not work, why does it not work when it was how we were taught in an Elementary Differential Equations course?

I can post the finding of the coefficients if need be, but since it was a few fairly ugly matrices, I left it out of the initial post.

Last edited:
transmini said:

## Homework Statement

Show that the steady state solution is given by $$x(t)=A(\omega)sin(\omega t-\phi)$$
3. The Attempt at a Solution
Which when combining the sine and cosine term into ##A(\omega)cos(\omega t+\phi)##
The problem statement asks for the solution in terms of a sine function with an argument that has ##\phi## subtracted. You found a solution in terms of a cosine function with an argument that has ##\phi## added.

I don't see why your method of solution would be considered incorrect.

TSny said:
The problem statement asks for the solution in terms of a sine function with an argument that has ##\phi## subtracted. You found a solution in terms of a cosine function with an argument that has ##\phi## added.

I don't see why your method of solution would be considered incorrect.
When I mentioned the combination of sine and cosine, the expression should actually be ##A(\omega)cos(]omega t-\phi)## rather than with ##+##. I've changed it in the original post. But because of that minus, I have a different phase shift, so it's not the EXACT answer the book was looking for. The solution the book gives has ##\phi## being positive, so it would shift the solution to the right, whereas since mine is negative, it is shifted to the left, thus not being the same solution to the equation of motion. This is why they call it incorrect. That was a typing mistake on my part.

I'm not understanding why you are putting your answer in terms of the cosine function while the problem asks for the answer in terms of the sine function.

TSny said:
I'm not understanding why you are putting your answer in terms of the cosine function while the problem asks for the answer in terms of the sine function.
I honestly have no idea. I just noticed that shortly before you posted this. I think maybe in our notes it's what we did so I just did that without thinking of the problem statement. I can't find verification other than one .com site, but combining into a sine term would be ##Rsin(\omega t +\phi)## correct? If so this would yield the correct answer since tangent and arctangent are odd, so the negative pulls out making it minus the positive ##\phi##.

You can convert ##A \cos \omega t + B \sin \omega t## into any of the following forms:

##A(\omega) \sin \left (\omega t + \phi \right)##
##A(\omega) \sin \left (\omega t - \phi \right)##
##A(\omega) \cos \left (\omega t + \phi \right)##
##A(\omega) \cos \left (\omega t - \phi \right)##

where, of course, the function ##A(\omega)## is different from the constant ##A##, and ##\phi## will be different for each of the four forms listed above.

Each form is fine. But your problem stated that it wants the form ##A(\omega) \sin \left (\omega t - \phi \right)##. If you express the answer in this form, then I think you should get expressions for ##A(\omega)## and ##\phi## as given in the problem statement.

TSny said:
You can convert ##A \cos \omega t + B \sin \omega t## into any of the following forms:

##A(\omega) \sin \left (\omega t + \phi \right)##
##A(\omega) \sin \left (\omega t - \phi \right)##
##A(\omega) \cos \left (\omega t + \phi \right)##
##A(\omega) \cos \left (\omega t - \phi \right)##

where, of course, the function ##A(\omega)## is different from the constant ##A##, and ##\phi## will be different for each of the four forms listed above.

Each form is fine. But your problem stated that it wants the form ##A(\omega) \sin \left (\omega t - \phi \right)##. If you express the answer in this form, then I think you should get expressions for ##A(\omega)## and ##\phi## as given in the problem statement.
If I used ##A(\omega)sin(\omega t+\phi)##, yeah. I'll have to mention how my answer is equivalent. I got a 10% on the problem with the work I have here. Thanks for helping spot the mistake I had missed.

transmini said:
If I used ##A(\omega)sin(\omega t+\phi)##, yeah. I'll have to mention how my answer is equivalent. I got a 10% on the problem with the work I have here. Thanks for helping spot the mistake I had missed.
OK. Note that the answer for ##\phi## will differ in sign for the two cases ##A(\omega)sin(\omega t+\phi)## and ##A(\omega)sin(\omega t - \phi)##.

So, it makes a difference that they asked for ##A(\omega)sin(\omega t - \phi)## rather than ##A(\omega)sin(\omega t + \phi)##

I definitely think you deserved more than 40%.

TSny said:
OK. Note that the answer for ##\phi## will differ in sign for the two cases ##A(\omega)sin(\omega t+\phi)## and ##A(\omega)sin(\omega t - \phi)##.

So, it makes a difference that they asked for ##A(\omega)sin(\omega t - \phi)## rather than ##A(\omega)sin(\omega t + \phi)##

I definitely think you deserved more than 40%.

I actually forgot to look at the back of the page, it turned out to be a 10%. Its all good though. Each homework assignment is just a tad over 1% of our overall grade so losing some points doesn't mean a whole lot. I will talk to my professor and see about getting a few back though. But yeah thanks again for helping see what was up with this problem.

OK. Good luck with retrieving some points.

## 1. What is the steady state solution of a forced, damped harmonic oscillator?

The steady state solution of a forced, damped harmonic oscillator is the solution that describes the behavior of the oscillator over time after it has reached a state of equilibrium. It is the solution that remains constant and does not change with time.

## 2. How is the steady state solution affected by damping?

The steady state solution is affected by damping because damping reduces the amplitude of the oscillations over time. This means that the steady state solution will have a smaller amplitude compared to the undamped solution.

## 3. What role does the forcing function play in the steady state solution?

The forcing function plays a crucial role in the steady state solution as it determines the frequency and amplitude of the oscillations. The steady state solution will have the same frequency as the forcing function and its amplitude will depend on the strength of the forcing function.

## 4. How is the steady state solution different from the transient solution?

The steady state solution is different from the transient solution because the transient solution describes the behavior of the oscillator over time until it reaches a state of equilibrium, while the steady state solution describes the behavior after the equilibrium has been reached. The transient solution will decrease over time, while the steady state solution will remain constant.

## 5. Can the steady state solution be used to predict the behavior of a forced, damped harmonic oscillator in the long term?

Yes, the steady state solution can be used to predict the long-term behavior of a forced, damped harmonic oscillator. This is because the steady state solution remains constant and does not change over time, making it a reliable prediction for the behavior of the oscillator in the long run.

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