Period is independent of amplitude. (Vias.org)

But given that,

Simple harmonic motion can be defined by

$$x = A * \sin(\omega t) \tag{1}$$ where $A$ is the amplitude of oscillation, $\omega$ the angular velocity, $t$ the time, and $x$ the displacement from the mean position


$$T = 2 \pi/\omega = 1/f \tag{2}$$ where, $T$ is the period of motion and $f$ is the frequency of oscillation.

Equation (1) can be rearranged to give \begin{align*} x &= A * \sin (\omega t)\\ \frac{x}{A} &= \sin(\omega ωt) \\ \arcsin\left(\frac{x}{A}\right) &= \omega t\\ \omega &=\frac{\arcsin\left(\frac{x}{A}\right)}{t} \end{align*}

Subbing this into (2) gives the following relationship between $T$ and $A$

$$T = \frac{2 \pi t}{\arcsin(x/A)} = \frac{1}{f}$$

Doesn't the fact that both $T$ and $A$ appear in the above equation show that $T$(period) is dependent on $A$(amplitude) ?


Although a physical explanation may be useful, I am particularity interested in why deriving a relationship between $T$ and $A$ doesn't mean that they are dependent. Note there is a similar question here but that is concerned with the physics of the phenomena, and not why the maths can't be used to solve it.

This is because I have this exam where we are given a stimulus and based on that stimulus alone are meant to answer questions (i.e. the exam is expected to contain material/principles that we haven't been exposed to before, but should be able to answer given the stimulus). And as I didn't know much about simple harmonic motion, my initial reaction was to see if the formulas link $T$ and $A$.

The practice question (which I have typed out the important bits of above is) Q Q The answer is D.

  • $\begingroup$ Just FYI, this site linked below can convert images to text e.g. from your image above: Simple harmonic motion can be defined by the equation = — k.x where F is the force acting and x is the displacement from the mean position. Alternatively, simple harmonic motion can be defined by x — A.sin wt where A is the amplitude and where the period of the motion T — 27r/co = 1 where f is the frequency of the simple harmonic motion. onlineocr.net $\endgroup$
    – user108787
    Commented Nov 5, 2016 at 0:34
  • $\begingroup$ Not sure if you're asking it this way but if you strike a guitar string hard the amplitude will be high as opposed to striking it soft and getting a low amplitude. However you strike it the frequency will be the same and is not dependent on the AmpliTube or vice versa. $\endgroup$ Commented Nov 5, 2016 at 0:37
  • $\begingroup$ Thanks for the link CountTo10, that will be very useful in future. @BillAlsept . Thanks for the explanation. Although that does help, I am more concerned about why the maths doesn't agree with the answer. The reason being, for this exam I am basing most of my answers (on different topics, not just simple harmonic motion) on what the maths shows me; but here the maths guides me to the wrong answer and am trying to find out why that is. I am assuming there is something different about the formulas which were provided which doesn't allow subbing one into the other. $\endgroup$
    – K-Feldspar
    Commented Nov 5, 2016 at 0:41
  • 2
    $\begingroup$ Don't you mean to say $T=\frac{2\pi}{\omega}=\frac{1}{f}$ $\endgroup$
    – M. Enns
    Commented Nov 5, 2016 at 0:47
  • $\begingroup$ Oh goodness, I think that is. The 2nd bar in the equals sign has probably disappeared (as you can see above) when the original page was scanned into a PDF which I have been reading off . Sorry! But thanks for picking that up. $\endgroup$
    – K-Feldspar
    Commented Nov 5, 2016 at 0:49

2 Answers 2


Just to drive home the point, your derivation does not in fact show period depends on amplitude. Let's stick with angular frequency(same deal). What we really have is $\omega(t) = \frac{\sin^{-1}(\frac{x(t)}{A})}{t} $. But now, how does x depend on t well, we then have $$\omega(t) = \frac{\sin^{-1}(\frac{A sin(\omega t )}{A})}{t} = \frac{\sin^{-1}(sin(\omega t))}{t} =\omega $$ In other words your derivation is a complete tautology

  • $\begingroup$ Thanks for explaining that. I don't know if there is an easy solution but is there any way to avoid doing that (like having a circular derivation or an x=x case) or to spot when that is happening? It didn't even occur to me that I might have done that. During exam time I won't really be able to check something over and over to make sure I haven't done that, so are there any tell tale signs that gave it away to you? $\endgroup$
    – K-Feldspar
    Commented Nov 5, 2016 at 1:09
  • 1
    $\begingroup$ All I can say is to use more than one equation to in derivations . That should reduce the chances of circular derivations $\endgroup$
    – Amara
    Commented Nov 5, 2016 at 3:23
  • $\begingroup$ Sorry I don't really understand what you mean. In the above derivation I used more than one equation. equation 1)x=A∗sin(ωt) and equation 2)T=2pi/w=1/f (which I wrote wrongly as T-2pi however). $\endgroup$
    – K-Feldspar
    Commented Nov 5, 2016 at 4:55
  • 1
    $\begingroup$ I should be more clear, it is good idea to use more than one independent equation the second equation is just the inverse of omega. So we can ignore the second and deal with just the first. So there is really one equation. $\endgroup$
    – Amara
    Commented Nov 5, 2016 at 5:29

It's because $x$ depends on $t$ in such a way that there is no dependence on $A$ left in the expression. $A$ and $\omega$ are constants that don't depend on time, and $x$ is a function of time; it's the position at time $t$, usually written $x(t)$.

When the force isn't Hooke's law, then you can get a relationship between $A$ and $\omega$. The reason for this is because the energy of a simple harmonic oscillator is $$ E = \frac{1}{2} m [v(t)]^2 + \frac{1}{2} k [x(t)]^2, $$ that is the equation for an ellipse in $x$-$v$ space no matter how big the amplitude/energy is. The simple harmonic osillator just rotates around that ellipse with a constant angular frequency, $\omega$. If I were to change the potential energy term to, say: $$ E = \frac{1}{2} m [v(t)]^2 + \frac{1}{4} k [x(t)]^4, $$ then the shape with a constant energy is no longer an ellipse, and the shape changes as the energy changes in such a way that amplitude and period become related.

For another example, a look at the orbits of planets and Kepler's third law — there amplitude along the major axis and period are related by: $$\frac{T^2}{A^3} = \frac{4\pi^2}{G(M+m)}$$


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