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Take this equation $$v^2 = 2 a \sin(2\pi b t)$$ where $v$ is a velocity and $t$ is a time. Find the dimensions of $a$ and $b$.

Finding the dimensions of $a$ is easy as the $\sin$ has no dimension so it will have the same dimensions as $v^2$.

But the problem is finding the dimensions of $b$ because the $\sin$ takes an angle as a parameter, and angles have no dimension we can simply say that $b$ should be $1/[\mathrm{T}]$ which will make the entire argument dimensionless. But isn't that just wrong?

Because there are some physical laws that have $\sin(t)$ for example where $t$ is a time, so it isn't a condition that the angle's value should be dimensionless.

This is really confusing.

N0va
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Ahmed Fwela
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  • why was this down-voted? – Ahmed Fwela Oct 05 '16 at 20:05
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    There are most certainly no physical laws with $\sin(t)$ in them where $t$ has the dimensions of a time. For $\sin$ to make sense its argument has to be dimensionless. See for example this Physics.SE question and its answers: http://physics.stackexchange.com/q/272599/ – N0va Oct 05 '16 at 20:09
  • You're correct, b would be in units of 1/t. Sines are simply ratios of lengths where the units cancel out. Perhaps the downvote was dues to "some physical laws that have sine(t)..." – bpedit Oct 05 '16 at 20:23
  • my teacher gave us some examples of dimensionally correct equations and one of them had sin(t) – Ahmed Fwela Oct 05 '16 at 20:25
  • Think of $sin (t) $ as short for $sin(t[t]\cdot 1 [t]^{-1})$, where $[t] $ is the units of t. Otherwise you'll have to finetune the coefficients to match the units of t you send in, as you do with b here. (Alternatively one decides beforehand what units are used. I.e. time is always in seconds, the angular frequency is always in Hz and so on) – Emil Oct 05 '16 at 20:52
  • Note that the series expansion of the sine function is: $$\sin x = x-\frac{x^3}{3!}+\frac{x^5}{5!}-...$$ For the dimensions of $x$, $x^3$, $x^5$, etc. to be consistent, it is required that $x$ is dimensionless. – nluigi Oct 05 '16 at 21:43
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    Think about a specific case, for example a particle performing simple harmonic motion with displacement $x = a \sin(2 \pi f t)$. Here $f$ is the frequency, with dimension (1/time) so $ft$ is dimensionless and so is $2\pi ft$. Of course you find problems in textbooks which say things like "the displacement of a particle is $10\sin(2t)$" but that is poor notation, with the excuse that the question is really about practising doing math, not about doing physics. – alephzero Oct 05 '16 at 21:53
  • @alephzero yea things like this in textbooks make me crazy – Ahmed Fwela Oct 05 '16 at 21:55

2 Answers2

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The arguments of trigonometric functions in physical equation have always to be dimensionless because the argument is a pure number. Therefore you are correct that the dimension of b has to be [1/T].

freecharly
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Inside the bracket of the sine function you have an angle which is dimensionless so the dimensions of $b$ will be $[T]^{-1}$.

If you did have a relationship with $\sin(t)$ in it then it means that inside the bracket you have $1\times t$ with the constant $1$ having the dimensions of $[T]^{-1}$.

Farcher
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  • but numbers have no dimensions right? – Ahmed Fwela Oct 05 '16 at 20:36
  • You asked about the constant$b$ so why cannot it be $1$ second$^{-1}$? – Farcher Oct 05 '16 at 20:57
  • on What basis should I take that? – Ahmed Fwela Oct 05 '16 at 20:58
  • On the basis that $b$ has the dimensions of $[T]^{-1}$ so I want $b$ to be numerically equal to $1$ or $2$ or $2 \pi$ etc – Farcher Oct 05 '16 at 21:00
  • my question is should the angle quantity represent a dimensionless value or not – Ahmed Fwela Oct 05 '16 at 21:02
  • An angle is dimensionless so $(t)$ must be dimensionless. So think of it as $(1 \times t) $ with dimensions $([T]^{-1} \times [T]) = (1)$ a dimensionless quantity. – Farcher Oct 05 '16 at 21:06
  • I think $\sin(t)$ really makes no sense, but on that discussion if it is meant as "$\sin(t*1[\mathrm{T}]^{-1})$" then one should really incorporate a unit for that $1[\mathrm{T}]^{-1}$ because if you do not do that the expression $\sin(t)$ could be anything really depending on what "kind" of $1$ you assume: 1/h, 1/s, 1/a. So $\sin(t/\mathrm{s})$ or $\sin(t/\mathrm{h})$ would be ok. $\sin(t)$ is just wrong because you did not specify the $1[\mathrm{T}]^{-1}$ there. – N0va Oct 05 '16 at 22:13
  • @M.J.Steil So what about something like $\sin(\omega t)$ with $\omega = 3 {\rm s}^{-1}$. Would you not write it as $\sin (3t)$? – Farcher Oct 05 '16 at 22:24
  • @Farcher No of course not I would write it as $\sin(\frac{3t}{\mathrm{s}})$ because if you want to plug in $\omega$ explicitly into an equation you have to keep its unit. If you plug in physical quantities into equations you can not just put in the magnitude and drop the unit. I mean if you have $x(t)=x_0+v t$ with $v=1\mathrm{m}/\mathrm{s}$ and $x_0=5\mathrm{m}$ you could write $x(t)=5\mathrm{m}+(1\mathrm{m}/\mathrm{s}) t$ but $x(t)=5+1t$ on the other hand is just non sense. – N0va Oct 05 '16 at 22:45