Wednesday 9 November 2016

Simple Harmonic Motion - What are the units for $omega_0$?


I'm trying to understand the units in:


$$mx''+kx=0$$



And the general solution is $$x(t)=A \cos(\omega_0 t)+B \sin(\omega_0 t).$$


Let $\omega_0 =\sqrt{\frac{k}{m}}$ - the unit for the spring constant $k$ is $kgms^{-2}$ or $Nm^{-1}$, where $m$ is in $kg$, so that the units of $\omega_0$ seem to be "per second" (i.e) $1/s$.


But, later we put $\omega_0$ in to the $cos$ and $sin$ functions which will return dimensionless ratios. So, The constants $A,B$ must be in $m$, since $x$ is in $m$.


What I don't understand is why my book says $\omega_0$ has the unit $rad/s$, I get that the input for cosine is $rad$ or some other angle measure, but where did the radians come from?


My analysis of units only proved $1/s$ as the actual units..!




I have just been informed that radians are dimensionless. So, that answers part of this question, yet I still don't know why we can't say that dimensionless one in degrees or rotations..? How do I know what kind of cosine and sine table to use with this dimensionless number?



Answer



Ah, good question. The radian is actually a "fake unit." What I mean by that is that the radian is defined as the ratio of distance around a circle (arclength) to the radius of a circle - in other words, it's a ratio of one distance to another distance. For an angle of one radian specifically, the arclength $s$ is equal to the radius $r$, so you get


$$1\text{ rad} = \frac{s}{r} = \frac{r}{r} = 1$$



The units of distance (meters or whatever) cancel out, and it turns out that "radian" is just a fancy name for 1!


Incidentally, this also implies that "degree" is just a fancy name for the number $\frac{\pi}{180}$, and "rotation" is just a fancy name for the number $2\pi$.


This actually addresses the edit to your question. Suppose that you had some object oscillating at $\omega = \pi/4\frac{\mathrm{rad}}{\mathrm{s}} = 0.785\frac{\mathrm{rad}}{\mathrm{s}}$, and you wanted to evaluate its position after 10 seconds. To get the cosine term, you would plug the numbers in, getting


$$\cos\bigl(0.785\tfrac{\mathrm{rad}}{\mathrm{s}}\times 10\mathrm{s}\bigr) = \cos(7.85\text{ rad}) = \cos(7.85)$$


and then you would go to a trig table in radians (or your calculator in radian mode) and look up 7.85.


However, suppose that you were measuring $\omega_0$ in degrees per second instead of radians per second. You would instead have


$$\cos(45^\circ/\mathrm{s}\times 10\mathrm{s}) = \cos(450^\circ)$$


If you go look this up in a trig table given in degrees, you will get the same answer as $\cos(7.85)$. Why? Well, remember that the unit "degree" is just code for $\pi/180$, so this is actually equal to


$$\cos\bigl(450\times\tfrac{\pi}{180}\bigr)$$


And $450\times\frac{\pi}{180} = 7.85$, which is just $450^\circ$ converted to radians. So now you have the same value in the cosine, $\cos(7.85)$. Trig tables listed in degrees already have this extra factor of $\frac{\pi}{180}$ built into them as a convenience for you; basically, if you look up any number $\theta$ in a table that uses degrees, what you get is actually the cosine (or sine, or whatever) of $\theta\times\frac{\pi}{180}$.



No comments:

Post a Comment

Understanding Stagnation point in pitot fluid

What is stagnation point in fluid mechanics. At the open end of the pitot tube the velocity of the fluid becomes zero.But that should result...