Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

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?

share|improve this question
    
See en.wikipedia.org/wiki/Radian_per_second –  ja72 Sep 21 '12 at 13:53
add comment

4 Answers 4

up vote 8 down vote accepted

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 $\theta\times\frac{\pi}{180}$.

share|improve this answer
    
I don't think this is right. 1 radian and 1 degree are both assigned the number 1, and both defined as the ratio of an arc length to radius, but using different standards for arc-length. The radian uses the radius, the degree uses the circumference divided up into 360 arc-lengths. Since angle = k arc-length/radius, then $k=180/\pi$ takes care of calculating 1 degree correctly for an arc-length = $2\pi r/360$ –  Physiks lover Aug 8 '12 at 21:58
    
@Physikslover I don't understand how you're claiming this isn't right. Surely you agree that $r/r = 1$, right? And also $r/r = 1\text{ rad}$ by definition. –  David Z Aug 26 '12 at 17:31
    
My previous comment wasn't quite correct and I should have said that angle is the ratio of arc-length to radius multiplied by a normalising factor K. You state: 'Incidentally, this also implies that "degree" is just a fancy name for the number π/180". I don't think this is correct because on one side you have the number 1(degree), on the other you have the ratio of two numbers(lengths) where the r's cancel to give $\pi/180$. so you need the conversion factor $k=180/pi$ in $1=k(2\pi r/360)/r$ –  Physiks lover Aug 26 '12 at 22:14
    
Sure... on the left you have the number 1, on the right you have the conversion factor $k$. So you multiply $1/k$ over to the left side and relabel it as a unit, namely the degree. That's how the degree is defined. –  David Z Aug 26 '12 at 23:51
    
which is equivalent to multiplying the lhs by (1/k)k so you're back to 1(degree) = (k= 180/π)(2πr/360)/r where every thing is consistent since the rhs = 1. But the rhs isn't just the ratio of two lengths, you still need k. –  Physiks lover Aug 27 '12 at 12:48
add comment

$\cos{(w_0t)}$ will be a solution of the given equation with $w_0=\sqrt{\frac{k}{m}}$ only if $\frac{d}{dx}\cos{x}=-\sin(x)$, which it will only be if $x$ is in units radians. If $x$ is in units degrees or grads there will be a dimensionless conversion constant, $\frac{d}{dx}\cos{x}=-C\sin(x)$. The convention that we use radians is so geometrically natural that it's close to hard-wired for mathematicians and physicists. I think the two Answers that appeared just as I started to write this are wrong in principle to take 1 radian=1, but in practice taking the radians unit to be dimensionless is unlikely to get you into trouble. Nonetheless, there's a reason why this is an international standard. Nice Question.

EDIT: I'm not entirely happy with the above. It depends whether we define $\cos{}$ to map an angle to a dimensionless ratio, or to map dimensionless numbers to a dimensionless ratio. The second definition would make the earlier Answers right and me wrong.

EDIT(2): Rather than writing $\sin{(60^o)}$ or $\sin{(\frac{60\pi}{180}^r)}$, we might write $\sin^{[o]}(60)$ or $\sin^{[r]}(\frac{60\pi}{180})$. When we use a table of sines, we check to see whether it's the sine-of-degrees function or the sine-of-radians function, then we give it the number $60$ or the number $\frac{60\pi}{180}$ as appropriate. There are different sine functions, which are related by linear transformations of their arguments. If we include translations, then $\cos$ becomes also a different sine function.

share|improve this answer
    
The point you make in your edit is quite right: this is a matter of definition. I'd claim that the second definition of (e.g.) cos is the "correct" (i.e., the most useful) one. That's because cos shows up in various contexts where the argument is not most naturally thought of as an angle. Think of $\cos(kx-\omega t)$ for a traveling wave, for instance. We can map $kx-\omega t$ onto an angle, if we want to insist that arguments of cos are always angles, but it's more natural to say that $kx-\omega t$ is just a dimensionless number, and cos takes arguments that are dimensionless numbers. –  Ted Bunn Jun 23 '11 at 19:43
    
@Ted, sine-of-radians is mathematically natural for many reasons, and sine-of-degrees is almost wholly only historically important, but sine-of-whole-rotations is relatively natural too. Factors of $2\pi$ move to different places in equations, of course. It's curious how much there is to this Question. It comes down to knowing how to recognize and translate between different conventions, as so often. –  Peter Morgan Jun 23 '11 at 21:28
    
<trig function> of radians gives the simplest mathematical form for a Taylor expansion, basically the same reason for taking exponentials and logs to the base $e$. –  dmckee Aug 25 '11 at 17:24
add comment

The importance of radians is "not" in its dimensions. It is necessary to mention whether $\omega_0$ is in radian/sec (and not just 1/sec) because angles can be expressed in various units - degrees, radians, etc (all which are dimensionless) and unless we know in which unit $\omega_0$ is expressed, we cannot do mathematical operations on it, like finding its sine or cosine, etc.
For example: $sin(60^o)$, where 60 is in degrees, is totally different from $sin(60^r)$ where 60 is in radians. So, it is "very" important to always mention $\omega_0$ in radians/sec rather than just 1/sec.

(This is the reason why $\omega_0$ is called as angular frequency and not just frequency, although both have the same dimensions.)

share|improve this answer
add comment

Radians are kind of a funny unit from the dimensional analysis perspective: radians are dimensionless. That means that rad/s and 1/s are equivalent from the point of view of dimensional analysis.

One way to think about this is that angular measures in radians are really just ratios of like quantities: $\theta$ in radians is, by definition, the ratio of the length of a circular arc subtending $\theta$ to the radius of the circle. So a radian is really a meter per meter.

In practice, when doing dimensional analysis in physics, this means that you can slip radians into and out of your units with wild abandon. For instance, if a circle of radius $r$ is rotating at angular speed $\omega$, then the speed of a point on the rim is $$ v=r\omega. $$ The right side of this expression has units of m rad/s, and the left side has units of m/s. But the units balance, because a radian (or a m/m if you prefer) is dimensionless.

share|improve this answer
    
+1, I like the comparison to meters per meter. –  David Z Jun 23 '11 at 17:23
    
I upvoted yours too. I like the observation that "degree" is just shorthand for $\pi/180$. As I'm sure you know, that fact is built into software packages like Mathematica: you can say things like Cos[60 Degree] and get the right answer, because Cos expects an argument in radians and Degree is defined to be that numerical value that does the correct conversion. –  Ted Bunn Jun 23 '11 at 17:26
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.