What Are the Units of b in a Damped Harmonic Oscillator Equation?

[BlackAntlers]

Homework Statement

Hello everyone, first time poster. I am in an Analytical Mechanics class and we are covering Dimensionless Form and Free Fall and am struggling with some of the concepts. The question has multiple parts, but looks like this. The equation for a damped harmonic oscillator is,

$m\ddot{x} = -kx + b\dot{x}$


a) What are the units of b?

b) What two variables will we need to make dimensionless by scaling?

c) For undamped harmonic motion what is the unit of time and how is it related to the spring constant K and the mass m?

d) Using this natural unit of time, construct the transformation that maps time to a dimensionless variable.

e) Using the time transformation you developed, turn the original equation into

${\frac{d^2 x}{dt^*2}} = -x + α {\frac{dx}{dt^*}}$

f) Identify α in terms of k, m and b. If you computed correctly, α should be dimensionless.

g)We have other constants available for scaling our diff eq. other than the ones that appear explicitly. We have two additional candidates for length and velocity, what are they and where do they come from?

Homework Equations

Period of a spring constant $T = 2π\sqrt\frac{m}{k}$

The Attempt at a Solution

a) We know mass times acceleration is a force F, whose units are N. N =$\frac {mass*length}{time^2}$. We know velocity $\dot{x} = \frac{length}{time}$ so, b must be $\frac{mass}{time}.$

b) Having an issue with this one because don't we want all variables that have dimensions to be dimensionless? Like mass, time, length and from those we could figure out acceleration and velocity in dimensionless terms?

c) The period for undamped harmonic motion is in seconds. The equation is in the relevant equations part. To figure out what K is, look in the original equation. X is a length, so K = $\frac{mass}{seconds^2}.$ When we cancel out our units in the period equation, we are left with seconds, which checks out. My only question is about the 2π. What do we do with this? Do we just ignore it?

d) Ok, so our $T = \sqrt\frac{M}{K} * T^*, (T^*)$ is our dimensionless unit for time.

Then $dT = \sqrt\frac{m}{k}*dT^*$ and $dT^2 = \frac{m}{k}*T^{2*}$

e) Dividing the mass out in original equation $\frac{d^2 x}{m/k}\frac{1}{dT^{2*}} = \frac{-k*x}{m} + \frac{b}{m} \frac{dx}{sqrt(m/k)}\frac{1}{dT^*}.$ So, distributing the $\frac{m}{k}$ and cleaning up we get ${\frac{d^2 x}{dt^*2}} = -x + α {\frac{dx}{dt^*}}$

f) I get $α = \frac{b}{m} \frac{m}{k}$ and when I plug in all the units they end up canceling out to be dimensionless, so that checks out.

g) This is the part I'm really stuck on. What are the two additional candidates for length and velocity? Do I have to add an $x_o$ and $v_o$ to the previous equation? And if I do, why exactly do I do this?

Thank you all in advance

 

[BvU]

Hello Ants, and welcome to PF. Hefty post !

Re b): you want dimensionless thingies so that you can take sines, logarithms, exponentials etc. We certainly don't want anything measurable to be dimensionless in physics, unless it's a ratio (and even then, mol/mol is different from kg/kg...)

Re c): I think the exercise wants you to claim the natural unit of time is the time needed to complete one full oscillation. The $2\pi$ is a ratio (length/length I would guess, but I'm on thin ice now...) and in view of the equation under d), apparently it's to be left out of the natural unit.

Re d): You really do want to write $dT^2 = \frac{m}{k}*dT^{*2}$, not $dT^2 = \frac{m}{k}*T^{2*}$ !

Re the damping: for some reason I am very used to $m\ddot{x} = -kx\; {\bf -} \;b\dot{x}$ with a positive k and a positive b, so that genuine damping takes place. But who am I?​

Re f): Could you show that ? I am stuck with a square root.

Re g): They've got me puzzled there too. Don't think they want velocity of light or such exotics. Have to pass for the moment. Someone else ?

PS what's this free fall in the title ?

 

[BlackAntlers]

BvU, thank you for your quick response.

b) I think I am missing some fundamental definition for this exercise. So, we have intensive variables like temperature, pressure and density, which are also our dimensionless variables. Then we have extensive variables like volume, force and mass which have dimensions. So we want to turn our spring equation dimensionless. Is that right so far? And in order to do so, we have to have all of our dimensions cancel with each other?

c) So our period is measured in seconds and seconds is our natural unit of time and 2 pi is some additional information not needed. Is it because it's a constant

d) Thanks for pointing that out, I wasn't sure. I'm just going by what the prof. wrote. He likes to make you think!

f) Yes, sorry about that piece of info. So, initially we have $\frac{b}{m} * \frac{m}{k}$*1/sqrt(m/k) But remember sqrt(m/k) is equal to our time in (s), so calculating it as a fraction gives us s^-1. So I just plugged our values into b and k and the dimensions cancel with each other.

g) The problem is asking for velocity and length that are hidden in this spring differential equation. But I'm blind as to where those come from.

pps the name from this came from our previous in class assignment, i guess he just didn't change the name X^O

 

[BvU]

Whoa, whoa: no such thing as dimensionless temperature, pressure, density! You cannot add a temperature and a pressure ! Check out this Si chart and the remainder of that site (well, it's a big site...).

Point is that for mathematical operations like taking the sine, a logarithm, etc. the argument of the function has to be dimensionless, so it's somewhat of interest to ensure that by a good choice of units.

Having said that, I must admit that the $2\pi$ threw me off a little bit, too (someone?).
The $\pi$ is very dimensionless, but more math than phys. Personally I have a hard time coming up with anything else than length/length, but I'm sure there are other ways of bringing it in. Really don't know what to do with the factor to go from cycle (period) to angle. The $\omega = 2\pi\;f$ doesn't have a preferred side as far as I can see.

In physics, there aren't many dimensionless constants. Fine structure constant $\alpha$ is the only one in the frequently used constants list here (and another choice of fundamental units stacks that up with a dimension too, I seem to remember).

I like your professor's approach: making people think for themselves is the best service one can do to the world as a whole.

And for g0: I can't for the life of me come up with a sensible scale for the amplitude. What's a natural extension for a spring? Or a natural angle for a pendulum ? All math and phys wants small oscillations to get easy equations. But 0 doesn't qualify for obvious reasons.

 

[paranormal]

for the help!

I can provide some guidance and clarification on the concepts you are struggling with. Dimensionless form and free fall are important concepts in analytical mechanics that help us understand the behavior of physical systems without being limited by specific units of measurement.

a) The units of b can be derived from the original equation m\ddot{x} = -kx + b\dot{x}. As you correctly stated, the units of mass times acceleration (m\ddot{x}) is equal to the units of force (N). Therefore, the units of b must be Ns/m, which can also be written as kg/s.

b) In order to make the equation dimensionless, we need to scale two variables. These variables are typically chosen to be the ones with units of length and time. In this case, we can scale x (length) and t (time).

c) The period of a spring constant (T) is related to the spring constant (k) and mass (m) by the equation T = 2π\sqrt\frac{m}{k}. This means that the period has units of seconds, and is related to the spring constant and mass in a specific way.

d) The transformation that maps time to a dimensionless variable can be written as t^* = \sqrt\frac{m}{k}t, where t^* is the dimensionless time variable. This transformation essentially scales the original time variable by the square root of the mass over the spring constant.

e) Using the time transformation, we can rewrite the original equation as \frac{d^2 x}{dt^{*2}} = -x + \alpha\frac{dx}{dt^*}. This is essentially the same equation as the original, but in a dimensionless form.

f) As you correctly stated, α = \frac{b}{m}\frac{m}{k}, which simplifies to just α = \frac{b}{k}. This means that α is a dimensionless quantity, as it should be in a dimensionless equation.

g) The two additional candidates for scaling are x and v. These are the position and velocity variables, respectively. By adding these to our equation, we can make it even more general and applicable to a wider range of systems. This is not always necessary, but it can be useful in some cases.

I hope this helps clarify some of the concepts you were struggling with