# Estimating damped harmonic oscillator parameters from this plot of the oscillations

• zenterix
zenterix
Homework Statement
Using a force of ##4\text{N}##, a damped harmonic oscillator is displaced from equilibrium by ##0.2\text{m}##.

At ##t=0## it is released from rest. The resultant displacement of the oscillator from equilibrium as a function of time is displayed below.
Relevant Equations
Using the given information, estimate as best you can

a) mass of oscillator

b) quality factor of oscillator
The equation of motion is

$$\ddot{x}+\gamma\dot{x}+\omega_0^2x=0\tag{1}$$

The roots of the characteristic polynomial are

$$\alpha=-\frac{\gamma}{2}\pm\sqrt{\frac{\gamma^2}{4}-\omega_0^2}\tag{2}$$

and since we have oscillations we are in the underdamped case in which

$$\gamma^2<\frac{\omega_0^2}{4}\tag{2a}$$

The solution to (1) is

$$x(t)=e^{-\frac{\gamma t}{2}}(c_1\cos{\omega t}+c_2\sin{\omega t})\tag{3}$$
where

$$\omega=\sqrt{\omega_0^2-\frac{\gamma^2}{4}}\tag{4}$$

The velocity is

$$\dot{x}(t)=-\frac{\gamma}{2}x(t)+e^{-\frac{\gamma t}{2}}(-c_1\omega\sin{\omega t}+c_2\omega\cos{\omega t})\tag{5}$$

Using the initial conditions ##x(0)=0.2## and ##\dot{x}(t)=0## we have

$$x(0)=c_1=0.2\tag{6}$$

$$\dot{x}(0)=-\frac{\gamma}{2}\cdot 0.2+c_2\omega=0\tag{7}$$

$$\implies c_2=0.1\frac{\gamma}{\omega}\tag{8}$$

Thus

$$x(t)=e^{-\frac{\gamma t}{2}}(0.2\cos{\omega t}+0.1\frac{\gamma}{\omega}\sin{\omega t})\tag{9}$$

$$\dot{x}(t)=-\frac{\gamma}{2}x(t)+e^{-\frac{\gamma t}{2}}(-0.2\omega\sin{\omega t}+0.1\gamma\cos{\omega t})\tag{10}$$

At this point, we've determined what the motion is given some unknown parameters.

Assuming the underdamping condition (2a) is true then each different choice of ##m, \gamma##, and ##k## determines the angular frequency of the resulting oscillations.

The forms of (9) and (10) don't make the amplitude and phase of the oscillations very explicit. These variable are, however, also a function of our choice of parameters.

We can see this by writing (3) in amplitude-phase form

$$x(t)=e^{-\frac{\gamma t}{2}} A\cos{(\omega t-\phi)}\tag{11}$$

$$A=\sqrt{c_1^2+c_2^2}=\frac{1}{10}\sqrt{4+\frac{\gamma^2}{\omega^2}}\tag{12}$$

$$\phi=\tan^{-1}{\frac{c_2}{c_1}}=\tan^{-1}{\frac{\gamma}{2\omega}}\tag{13}$$

$$\dot{x}(t)=-\frac{\gamma}{2}x(t)-e^{-\frac{\gamma t}{2}}A\omega\sin{(\omega t-\phi)}\tag{14}$$

At this point, we look at the given plot

My question is about estimating the parameters from this plot.

Recall that what we want to determine ultimately is the mass that is oscillating and the quality factor of the oscillator.

The latter is defined by

$$Q=\frac{1}{\gamma}\sqrt{\omega_0^2-\frac{\gamma^2}{4}}\tag{15}$$

Here is what I did.

The first thing I did was to determine the spring constant from the original differential equation.

$$\ddot{x}(0)+\gamma\dot{x}(0)+\omega_0^2x(0)=0\tag{16}$$

$$\frac{F_s}{m}+\omega_0^2\cdot 0.2=0\tag{17}$$

where at ##t=0## the magnitude of the spring force equals the force we were applying to hold the spring in its initial position, and the sign is opposite.

$$\frac{-4}{m}+\frac{k}{m}\cdot 0.2=0\tag{18}$$

$$\implies k=20\mathrm{\frac{N}{m}}\tag{19}$$

In words, we determined the spring constant by noting that at position ##0.2## the spring exerts a force of ##-4\text{N}## and this force has the formula ##F_s=-kx##. Thus, ##k=\frac{F_s}{x}##.

Suppose we try to estimate the pseudo-period from the plot.

I've drawn in some dashed lines in the plot to try to do this

The pseudo-period seems to be slightly larger than 3. Perhaps 3.2.

Before I continue, I have a question.

Since the mass is released from rest, can we assume that the phase is zero?

From (13) it seems we cannot as this would imply that ##\gamma=0##. But if we are holding the mass and releasing, doesn't it start at the maximum of a pseudo-oscillation? Ie, at the top of the damped cosine? Is it not then in phase with cosine?

WWGD
Even though the issue with the phase is confusing to me, let me show another path that doesn't rely on assuming anything about the phase.

So far we know that ##k=20##.

We have two unknown parameters ##\gamma## and ##\omega_0##.

##\omega##, ##A##, ##\phi##, and ##m## are all of interest but are functions of the two parameters.

Note that we have position and velocity as functions of time, but given time these are functions of the two parameters as well.

We also have the period ##T## which is a function of ##\omega## and thus of the two parameters.

If we assume (based on the plot) a value for any two of the variables that are functions of the two parameters we will have two equations in the two unknown parameters.

Here is an attempt to do this.

First, let's take a guess at what the (pseudo) period is. It looks like it is slightly more than 3. Let's guess

$$T\approx 3.2$$

This gives us

$$\omega=\frac{2\pi}{T}=1.96$$

Next, it looks like position is about 0.18 at time ##T##.

$$x(T)\approx 0.18$$

$$x(T)=e^{-\gamma T/2}A\cos{(2\pi-\phi)}=0.18$$

$$e^{-\gamma T/2}A\cos{\phi}=0.18$$

And using ##T=3.2## together with ##A\cos{\phi}=0.2## from previously we can solve for ##\gamma##

$$\gamma\approx 0.065$$

At this point we can obtain all the other variables.

$$\omega_0\approx 1.96$$

$$m\approx 5.18$$

$$A\approx 0.200028$$

$$\phi \approx 0.01676$$

The quality factor is

$$Q\approx 29.81$$

These results, namely a mass of about 5kg and a quality factor of about 30 are in line with the answer to this problem (which is from MIT OCW and can be seen here).

The question I have is that it does seem that ##\phi## should be exactly zero. Should it be zero?

Why could I not have started with the assumption that it is zero and gone from there?

If we assume ##\phi=0## from the start we get

$$\tan{\phi}=\frac{\gamma}{2\omega}=0$$

$$\implies \gamma=0$$

But if ##\gamma=0## then there is no damping. So this can't be the case.

Here are my thoughts on my question.

Consider ##\cos{t}## and compare this to ##e^{-0.2 t}\cos{t}##.

The red graph is the usual cosine. The blue is the damped cosine.

What we see is that the damped cosine with no phase lag in the cosine appears, visually, to be lagging the normal cosine.

If we consider region of the graphs near ##t=0## what we have for the blue graph is that below 0 the cosine is slowly decreasing but the exponential is increasing rapidly. If we get more negative, the cosine eventually wins and so the blue graph starts decreasing.

The blue graph is what would model the mass in the OP. Since we know that the velocity is zero at ##t=0## we need to have the maximum of the blue graph occur at ##t=0## and so it must have a non-zero phase lag. In fact, this phase lag is going to be negative.

zenterix said:
Consider ##\cos{t}## and compare this to ##e^{-0.2 t}\cos{t}##.
Damping changes the frequency (slightly) so the equations you should be comparing are:
##y = \cos({\omega_0 t})## and
##y = e^{-0.2 t}\cos({\omega_D t})##

A few other points...

Remember the question asks you to ‘estimate’ – which suggests reasonable approximations can (should?) be made.

To find the period as accurately as possible, note that 19 oscillations occur in 60s. This gives ##T \approx 3.16## s. Don’t just use a couple of oscillations.

For reasonably lightly damped oscillations (as is the case here) a sensible approximation is ##\omega_D \approx \omega_0##. Using this makes finding ##m## simple.

If the decay factor is ##e^{{-\frac t{\tau}}}## then I believe a handy formula is ##Q = \dfrac {\tau \omega_0}2##. (To find ##\tau## use the graph to find the time taken for the amplitude to fall to ##\frac 1e## of its initial value.)

Avoid inappropriate use of significant figures, e.g. ‘##Q\approx 29.81##’ is an unjustifiable precision.

zenterix
If you assume $x(t) = Ae^{-\alpha t} \cos \beta t$ then you will find $$\dot x(t) = Ae^{-\alpha t}(-\alpha \cos \beta t - \beta \sin \beta t)$$ so that if $\dot x(0) = 0$ then $A \alpha = 0$, which is a contradiction. If instead $x(t) = Ae^{-\alpha t}\cos \beta (t-t_0)$ then $$\dot x(t) = Ae^{-\alpha t} ( -\alpha \cos \beta(t - t_0) - \beta \sin \beta(t - t_0) ).$$ Thus if $\dot x(0) = 0$ it must follow that $$\alpha \cos \beta t_0 - \beta \sin \beta t_0 = 0$$ which implies $$t_0 = \frac 1 \beta \arctan\left( \frac \alpha \beta \right).$$

$\alpha$ and $\beta$ can be directly estimated from the graph by considering the loss of amplitude over a period and the length of a period respectively. Then $\gamma$ and $\omega_0$ can be determined from $\alpha$ and $\beta$.

At $t = 0$ we must have $$4\,\mathrm{N} = m|\ddot x(0)|$$ and we can easily find $\ddot x(0)$ in terms of $\alpha$ and $\beta$ by differentiation.

zenterix
Steve4Physics said:
Damping changes the frequency (slightly) so the equations you should be comparing are:
When we consider a function (e.g. the solution ##x(t)##) that has a cosine in it that has an undetermined phase, the phase represents an angular lag relative to the exact cosine without the phase, right?

This is the comparison I made in post #3 above.

zenterix said:
When we consider a function (e.g. the solution ##x(t)##) that has a cosine in it that has an undetermined phase, the phase represents an angular lag relative to the exact cosine without the phase, right?
.Consider ##u(t) = \cos t## and ##v(t) = e^{-0.2t}\\cos t##.

The zeroes of ##u(t)## and ##v(t)## occur at the same values of ##t##, i.e. at ##t = \frac {\pi}2, \frac {3\pi}2##, etc.).

The extrema (maxima and minima) for ##u(t)## occur at ##t =0, \pi, 2\pi,## etc.). But the extrema for ##v(t)## occur at slightly different values of ##t##.

That means the shape of ##v(t)## is based on a ‘distorted’ cosine.

If you look at a more extreme (heavier damping) case this might help illustrate what’s happening. E.g. compare the graphs for ##\cos t## and ##e^{-t}\cos t##.

Last edited:
zenterix and hutchphd
zenterix said:
Homework Statement: Using a force of ##4\text{N}##, a damped harmonic oscillator is displaced from equilibrium by ##0.2\text{m}##.

At ##t=0## it is released from rest. The resultant displacement of the oscillator from equilibrium as a function of time is displayed below.
Relevant Equations: Using the given information, estimate as best you can

a) mass of oscillator

b) quality factor of oscillator

Before I continue, I have a question.

Since the mass is released from rest, can we assume that the phase is zero?
The phase is a constant fixed by initial conditions and not a parameter associated with the physical system. Having defined the form of the solution, it is up to you to determine the meaning of ##\phi##....no assumption is required other than meeting the initial conditions. I think you do need to worry about dividing by zero as you try to extract blood from this particular turnip. Also why bother??. Again the system "Q" is not dependent upon initial conditions and for a hi Q oscillator is best estimated by counting the oscillations over a half life or perhaps a decay (1/e) time.

zenterix and Steve4Physics
pasmith said:
$\alpha$ and $\beta$ can be directly estimated from the graph by considering the loss of amplitude over a period and the length of a period respectively.
Finding the loss of amplitude is not quite straightforward because of the contributions from the trig factor and the exponential factor. Should one measure the amplitude at the apparent extrema or half way between the x axis intercepts? Either works, but the algebra to extract ##\alpha## is a little different.

zenterix and hutchphd
hutchphd said:
Again the system "Q" is not dependent upon initial conditions and for a hi Q oscillator is best estimated by counting the oscillations over a half life or perhaps a decay (1/e) time.
That's a neat/simple strategy. I hadn't heard of it before but on searching I found this:

"The ring-down method is extremely simple: first you have to "shake" the resonator someway to make it oscillate at its natural frequency. Then, while observing the amplitude of the oscillations decrease, count how many cycles it takes to halve the amplitude and multiply this number by 4.53 to find Q."
(from https://www.giangrandi.ch/electronics/ringdownq/ringdownq.shtml )

As explained in the link, the number 4.53 is ##\frac {\pi}{ln(2)}##.

zenterix and hutchphd

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