The equation of motion of a particle is $x = A \, \mathrm{cos}\left[(\alpha t)^2\right]$. What type of motion is it?

The answer to this question in my textbook was: "Oscillatory but not periodic". How is this possible? The first line of my textbook says: "Every oscillatory motion is periodic but not every periodic motion is oscillatory." So, is the answer provided incorrect, or is my textbook incorrect?

I think this has something to do with time being a quadratic function, but I can't seem to think any further. This isn't one of those usual SHM equations you see.

Question: Is the motion $x = A \, \mathrm{cos}\left[(\alpha t)^2\right]$ oscillatory, periodic, both or neither?

  • 22
    $\begingroup$ I think that, without a clear definition of what "oscillatory" means, none of this makes any sense. (Your textbook's fault, not yours.) $\endgroup$
    – d_b
    Apr 13 at 16:48
  • $\begingroup$ @d_b In my textbook, "oscillatory motion" is defined as any motion that moves back and forth about a fixed point after a regular interval of time. However, I don't find this definition to be very helpful, although I could be mistaken. $\endgroup$
    – Haider
    Apr 13 at 16:57
  • 22
    $\begingroup$ I second the idea that this textbook is just flat out self contradictory and crap. $\endgroup$ Apr 13 at 17:11
  • 18
    $\begingroup$ Not to contradict, but I think a student being rightfully confused by their textbook’s limitations and attempting to properly understand the concepts is a perfectly valid use of this site. It is not high level, but valuable for new learners. $\endgroup$ Apr 13 at 17:18
  • 1
    $\begingroup$ @Krishnadev N, your recent edit changed the letters and types of parentheses used in this question. When I edited, I intentionally refrained from doing this as I assumed that the OP's original version would hew closely to the actual formula found in the textbook in question. Changing the typography changes the subject of the question, as the number of answers addressing the typography makes clear. $\endgroup$
    – tobi_s
    Apr 16 at 2:39

8 Answers 8


enter image description here

The graph of $cos^2(t)$ (blue) oscillates and is periodic in time.

The graph of $cos(t^2)$ (green) oscillates but is not periodic. The time between cycles is always decreasing

The answer to this question in my textbook was: "Oscillatory but not periodic". How is this possible? The first line of my textbook says: "Every oscillatory motion is periodic but not every periodic motion is oscillatory". So, is the answer provided incorrect, or is my textbook incorrect?

The answer provided by your textbook is correct for $cos(t^2)$ but the first line is wrong and should read:

Oscillatory motion can be either periodic or aperiodic with respect to time.

  • 2
    $\begingroup$ "The time between cycles is always decreasing"... for $t > 0$. $\endgroup$ Apr 14 at 4:31

The formula is ambiguous! $$ \cos( (\alpha t)^2 ) $$ oscillates but is not periodic.

$$ ( \cos(\alpha t) )^2 $$ oscillates and is periodic.

  • 8
    $\begingroup$ It is even worse: $acos$ is sometimes used for $arccos$ ;) $\endgroup$
    – Semoi
    Apr 13 at 18:15
  • 2
    $\begingroup$ And it goes without saying that no physical object could oscillate according to the first formula indefinitely. $\endgroup$
    – EvilSnack
    Apr 14 at 2:25
  • 12
    $\begingroup$ @Semoi actually, $\operatorname{acos}(x)$ is always the same as $\operatorname{arccos}(x)$—unlike $a\cos(x)$, which is a product. Italics and spacing do make a difference. $\endgroup$
    – Ruslan
    Apr 14 at 12:19
  • $\begingroup$ Actually there is no ambiguity, at least not as soon as the formula is cleaned up to make clear the operator being applied is $\cos$ (which was not so in the very first version of the question): there are precedence rules for operators like $\lim$ and $\exp$ and $\cos$, even though they are not taught at school when things like addition and multiplication are treated. The rule is that these operators, just like $\sum$ and $\int$, all have the same binding power, which is weaker than multiplication and division, but stronger than addition and subtraction. So $\cos 3x^2+1$ is $(\cos(3x^2))+1$ $\endgroup$ Apr 16 at 9:35
  • $\begingroup$ @MarcvanLeeuwen I agree there are precedence rules but only some of them are universally agreed and applied. You will not see an expression such as $\cos(at)^2$ in most science and maths texts, for reasons of clarity. $\endgroup$ Apr 16 at 15:03

If the expression is intended to be $x = a (\cos (\alpha t))^2$ then this is most definitely periodic, with period $\frac {\pi}{\alpha}$. In other words:

$x(t) = x(t \pm \frac {\pi} {\alpha}) = x(t \pm \frac {2 \pi} {\alpha}) =x(t \pm \frac {3 \pi} {\alpha}) \dots$

If the expression is intended to be $x = a \cos ((\alpha t))^2)$ then you could argue that this is oscillatory but not periodic - but this depends on your definition of oscillatory.

The answer "oscillatory but not periodic" seems to contradict the claim in your textbook that "Every oscillatory motion is periodic". The usual definition of "oscillatory" is a synonym for "periodic" - with this definition any motion that is oscillatory is also periodic and vice versa. If you adopt a broader definition of "oscillatory" to mean any motion or function that is bounded between two extreme values, then it is possible to have motions that are oscillatory but not periodic (e.g., a random walk that is bounded by $\pm 1$), but I cannot think of any definition that would allow a motion to be periodic, but not oscillatory.

  • 3
    $\begingroup$ The period is $\pi$ not $2\pi$ since it is squared $\endgroup$
    – LolloBoldo
    Apr 13 at 17:14
  • 1
    $\begingroup$ Maybe some breath before answering and accepting could help. If the square is on the cos function, the period is $\frac{\pi}{\alpha}$. $\endgroup$ Apr 13 at 17:30
  • $\begingroup$ Agreed. I have amended my answer. $\endgroup$
    – gandalf61
    Apr 13 at 17:49
  • $\begingroup$ @gandalf61 How about a Dirac delta function that occurs every fixed interval? $\endgroup$
    – Rich
    Apr 13 at 18:20
  • 1
    $\begingroup$ @StevanV.Saban In your example the velocity $v_x$ is periodic and oscillatory but the displacement $x$ is neither periodic nor oscillatory. $\endgroup$
    – gandalf61
    Apr 14 at 6:49

I agree with others in that the "definition" (if you can call it that) of oscillatory in your book leaves something to be desired, and damped oscillation as suggested by LolloBoldo is still oscillation in my books.

As a math guy (caveats right there) I would think that a linear combination of two oscillatory functions is still oscillatory (but I refuse to give a precise definition of oscillatory, because I think the word is descriptive only). The same does not hold for periodic functions unless the periods are integer multiples of a common period. A standard example of that is the wave

$$f(x)=\cos x+\sin(\sqrt2 x).$$

Enter image description here

This is not periodic, because $\sqrt2$ is irrational. I would still say that such waves oscillate, but that is my opinion and may be against mainstream use of the word. Observe that such waves can be physical (as opposed to just a mathematical concoction). We can have a system with two modes of vibration with the ratio of the (eigen)frequencies an irrational number. Then most solutions are superpositions of pure ($\approx$ periodic) waves, and share some characteristics of the function shown above.

The Bessel functions brought up by P-Dub are another example of oscillatory, but nonperiodic, vibrations that are physical.


Periodic usually means that:

$$x(t+ T) = x(t)$$ Where $T$ is the period. Based on that clearly $x(t) =K\cos(\alpha t)^2$ is periodic of period $T= \pi/\alpha$.

If one assume that oscillatory means that fixed a central value $V$ the function goes up and down around $V$ in a periodic fashion, i.e. you oscillate around $V$ every $T$, then you can have oscillations which are non periodic.

An example is the damped oscillator: since the amplitude depends on $t$ you don't have $x(t) = x(t+T)$ anymore, but nonetheless you have oscillations, since the $x(t)-V$ switches every period $T$.

You case falls under both: is periodic AND oscillatory


I would question the notion that the usual definition of "oscillatory" implies periodicity. Otherwise the term "chaotic oscillation" does not make sense, but indeed it is used a fair lot.

Non-periodic oscillatory behaviour is in fact a typical characteristic of chaotic systems. The Lorenz system of differential equations is one of the most well known examples with the iconic butterfly shaped Lorenz attractor. See Lorenz System, but also the more general Chaos Theory on Wikipedia.

The more specific aspect of your question regarding the stated "equation of motion" has been dealt with detail in the other answers. But note that this is not an equation of motion, but the solution to one. An equation of motion is a differential equation in the dynamic variables.


I would formalise oscillatory and periodic as follows: Take a function $f: \mathbb{R} \to Y$, where $\mathbb{R}$ are the time-values and $Y$ are the possible values of $f$ (most often position, amplitude or some other physical quantity), then

  • $f$ is periodic if there is a $T > 0$ such that for every $t \in \mathbb{R}$ it holds that $f(t + T) = f(t)$. The smallest possible choice of $T$ is then called the period of $f$.
  • $f$ is called oscillatory if there is a $y \in Y$ such that $f$ obtains $y$ infinitely often, i.e. for every time $s \in \mathbb{R}$ there are other times $t > s > r$ such that $f(t) = y = f(r)$.

Then as in the answer of @Stevan V. Saban $t \mapsto \cos(t)^2$ will be periodic and oscillatory, but $t \mapsto \cos(t^2)$ will be oscillatory but not periodic. The period $T$ of the first one will be $\pi$, and it e.g. obtains $y = 1$ infinitely often, but the second one does not have a period but still obtains $y = 1$ infinitely often.


For example, Bessel functions are oscillatory but not periodic.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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