I'm trying to understand the units in:


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?


8 Answers 8


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


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

  • $\begingroup$ 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 = $2\pi r/360$ $\endgroup$ Aug 8, 2012 at 21:58
  • $\begingroup$ @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. $\endgroup$
    – David Z
    Aug 26, 2012 at 17:31
  • $\begingroup$ 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 $1=k(2\pi r/360)/r$ $\endgroup$ Aug 26, 2012 at 22:14
  • $\begingroup$ 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. $\endgroup$
    – David Z
    Aug 26, 2012 at 23:51
  • $\begingroup$ 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. $\endgroup$ Aug 27, 2012 at 12:48

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.

  • $\begingroup$ 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. $\endgroup$
    – Ted Bunn
    Jun 23, 2011 at 17:26

In addition to all the above extremely well written answers, especially by David Z and Ted Bunn, let me tell you how you can visualize the origin of the $rad/s$ in your dimensions for $\omega_0$.

First, we note that a simple harmonic motion is similar to a uniform motion on a circle. To see how, we will work with a specialise case where the particle is pulled to a distance $r$ away from the mean position and released and we start our clock when it is passing through its mean position i.e. $x=0$ when $t=0$. Thus we get the equation of motion as $x=r sin(\omega t)$.

Next we draw a circle with radius $r$ and centre at the origin, and place the particle at $(r,0)$. We then let the particle move at a constant angular velocity $\omega$.

circular motion

If we now look at the parametric coordinates of the particle when it is at an angle $\theta$ from the X-axis, we get its current position. Concentrating on only the Y-coordinate, we see that it corresponds to $y=r sin\theta$, or in terms of $\omega$, we have $y=r sin(\omega t)$. Hmm... looks almost like the equation of the second paragraph. Infact, if we look at another particle that can only move along the Y-axis and coupled to this particle, it actually does the exact simple harmonic motion we had with our spring system. Thus, uniform circular motion is exactly analogous to simple harmonic motion, and applying this analogy to our equations, we see that $\omega$ can be defined as the angular velocity of the particle and ths has dimensions of $rad/sec$ (since $\omega$ is by definition $d\theta/dt$).

But when doing physics, its actually better to have the arguments of the trigonomtric functions as dimensionless. Why? A look at the Taylor series of the $cos$ function might help :


In the RHS, the first term $1$ is dimensionless. By the property of dimensional analysis, all additive terms must be of the same dimensions and thus have to be dimensionless. Thus, $x$ has to be dimensionless, and it is the argument of the cosine function. Hence, the requirement.

PS: There was a time when physicists actually let the angular frequency $\omega$ have the dimensions of $rad/sec$, to differentiate it from normal frequency $\nu$ or $f$ with a dimension of $1/sec$. But, nowadays its more natural to use $\omega$ for all frequency purposes because it fits more naturally into quantum mechanics, fourier transforms, special relativity etc.


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.)


$\cos{(w_0t)}$ 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.

  • $\begingroup$ 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. $\endgroup$
    – Ted Bunn
    Jun 23, 2011 at 19:43
  • $\begingroup$ @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. $\endgroup$ Jun 23, 2011 at 21:28
  • $\begingroup$ <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 $e$. $\endgroup$ Aug 25, 2011 at 17:24
  • $\begingroup$ Yes, the $\sin$ that takes inputs with units degrees and the $\sin$ that takes unitless inputs are two different functions. $\endgroup$
    – ryang
    Feb 6, 2021 at 18:20

I find it helpful to use rad in units to understand these mechanical properties:

The unit for spring constant k for a torque spring (clockwork spring) is N.m, but call it (N.m)/rad and it is easier to understand: "Torque per angle".

Also N.m is "formally" the same as J, and this can make it hard to distinguish from torque, especially when the work is associated with a rotational movement. It helps then to think of Joule as N.m.rad, meaning "torque times angle turned" analog to the way work from linear motion is "Force times distance traveled".

I would never call Joule anything else than J in documentation, it just helps to think about it that way.

Just a thought, rad is of course "1" and can be left out at will. I do not see how leaving out rad would lead to confusion with degrees. In engineering, degrees is used as a unit all the time, but never without the little degree sign °


Radians are a unit of Angle, in a trigonomentric sense, which brings repetition with rotation (cycles & revolutions). There are other units of Angle, such as the degree.

SI defined the radian to be it's Base Unit, and then gave it the name "Supplementary Unit" to get out of an impasse between factions. In those days calculations were often done with pen and paper, and dimensional analysis was split out (the Calculus) and done separately. Some have even suggested you can take the tangent of 12 inches! (so says the Calculus).

The issue is that the Base Unit of the metre is not a dimension. Rather it is a measure in a 3d space.

This means that you can divide one dimension by another different dimension (e.g. height, the direction of gravity, by the width) and get an apparently dimensionless number.

The "Angle Dimension" concept is simply a way of recording that we previously had two dimensions (Lx, Ly), and now we (would) have none.

We would never allow Temeperature / Time to be cancelled just because they have the initial letter T.

Many modern computer algebra systems (Mathcad, Mathematica, Maple) do cope with the auto conversion between units and check the resulting dimensions, with scaling, to support many science and engineering calculations. However all are stymied by Torque vs Work because the Angle unit is missing. The Torque problem extends into mechanical CAD systems as well.

The ability to label and check the apparently "dimensionless" numbers is a sad loss.

Mach 3 + 4 radians = 7 Reynolds.

Also note that there are no named dimensions at all in mathematics, so don't let them tell you that oft repeated mantra "Obviously all angles are in radians" without a little bit of thought and challenge.

E.g. Look up the Cordic algorithm, which is defined on the eight of a turn (45 degrees) where tan(45deg)=1, what a convenience! (i.e. the regular sin(x)=x+.. is just another convenience!)


This answer is largely an add-on to Yuzuriha Inori's.

You're right, the dimensional analysis of $\omega_0$ does work out to $1/s$. The inclusion of $\mathrm{rad}$ cannot be explained via dimensional analysis. The inclusion of $\mathrm{rad}$ happens when you transform/map/analogize the motion of the mass from a line to a circle. This transformation/mapping/analogy is the subject of Yuzuriha Inori's answer.

Consider a $1\,\mathrm{kg}$ mass pulled by a $1\,\mathrm{N}\cdot\mathrm{m}$ spring starting at $1\,\mathrm{m}$ of stretch. Once released, it travels to $-1\,\mathrm{m}$. In the physical spatial domain, its trajectory is a line. It starts off slow, accelerates through $0\,\mathrm{m}$, then slows down. As it turns out, it will take $\pi$ seconds to get to $-1\,\mathrm{m}$.

Mapping this linear back-and-forth trajectory from the physical domain to the time domain, the trajectory becomes a cosine curve with a half-period of $\pi$ seconds. In turn, mapping this cosine curve to the circle that Yuzuriha Inori describes, the trajectory is a now half-circle arc with subtended angle $\pi\,\mathrm{rad}$. This is where the $\mathrm{rad}$ comes from: the mapping to an angles domain. There are no angles anywhere in the physical domain, nor are there any when you map to the time domain, but when you map to a circle now you have angles and $\mathrm{rad}$. Yuzuriha did not get into this following point and neither will I, but just making a note of this here for interest, the circle is the mapping of the mass's trajectory in the complex domain. Thinking of this mass moving back and forth along a line in the physical domain abstractly as an object moving around a circle in the complex domain ends up useful for all kinds of analysis.

Your question of where the $\mathrm{rad}$ comes from has an underlying deeper question: why does a $1\,\mathrm{kg}$ mass pulled by a $1\,\mathrm{N}\cdot\mathrm{m}$ spring starting at $1\,\mathrm{m}$ of stretch happen to take $\pi$ seconds to reach $-1\,\mathrm{m}$? Sure, there are many ways to prove that it does, but how to explain why? The fact that it does enables this mapping to a sinusoid and then a circle and subsequent analysis. But the underlying why is not something I know how explain, any more than I can explain why $e^{i\pi}+1=0$.


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