While in my physics classes, I've always heard that the simple harmonic motion formulas are inaccurate e.g. In a pendulum, we should use them only when the angles are small; in springs, only when the change of space is small. As far as I know, SHM came from the differential equations of Hooke's law - so, using calculus, it should be really accurate. But why it isn't?

  • 17
    $\begingroup$ Hooke's law is an idealisation. $\endgroup$ Commented May 29, 2016 at 14:51
  • 1
    $\begingroup$ cyberphysics.co.uk/graphics/graphs/stress_strain2.gif $\endgroup$
    – valerio
    Commented May 29, 2016 at 14:54
  • $\begingroup$ related: Why does $\omega = \sqrt{V''(x_0) / m}$? $\endgroup$ Commented May 29, 2016 at 15:00
  • $\begingroup$ SHM implies the motion continues unabated, forever. But it doesn't. Thus SHM is an idealization. $\endgroup$ Commented May 29, 2016 at 23:38
  • $\begingroup$ Sometimes, though, Hooke's law is a very good idealization. For the pendulum, one can clearly see that $\ddot{\theta} = -\omega^2\,\sin\theta$ is a better description of the physics. But for a spring working under small strain, Hooke's law is remarkably accurate (the most significant departure from ideal is often the loss, which doesn't need to be a nonlinear effect) and a more accurate nonlinear equation is not needfully $\ddot{\theta} = -\omega^2\,\sin\theta$ but some more general nonlinear function that approximates $\theta$. $\endgroup$ Commented May 30, 2016 at 6:32

7 Answers 7


The actual restoring force in a simple pendulum is not proportional to the angle, but to the sine of the angle (i.e. angular acceleration is equal to $-\frac{g\sin(\theta)}{l}$, not $-\frac{g~\theta}{l}$ ). The actual solution to the differential equation for the pendulum is

$$\theta (t)= 2\ \mathrm{am}\left(\frac{\sqrt{2 g+l c_1} \left(t+c_2\right)}{2 \sqrt{l}}\bigg|\frac{4g}{2 g+l c_1}\right)$$

Where $c_1$ is the initial angular velocity and $c_2$ is the initial angle. The term following the vertical line is the parameter of the Jacobi amplitude function $\mathrm{am}$, which is a kind of elliptic integral.

This is quite different from the customary simplified solution


The small angle approximation is only valid to a first order approximation (by Taylor expansion $\sin(\theta)=\theta-\frac{\theta^3}{3!} + O(\theta^5)$).

And Hooke's Law itself is inaccurate for large displacements of a spring, which can cause the spring to break or bend.

  • 9
    $\begingroup$ Why are $\text{a}$ and $\text{m}$ written in non-math fonts in the equation? Why is the formula for $\theta(t)$ not an oscillatory function? $\endgroup$
    – DanielSank
    Commented May 29, 2016 at 15:38
  • 4
    $\begingroup$ In addition to David Sanks' question, what are $c_1$ and $c_2$ and what does $|$ stand for? $\endgroup$
    – Gert
    Commented May 29, 2016 at 15:49
  • 1
    $\begingroup$ The solution is from Mathematica, and was converted to TeX using the TeXForm command. I will correct the formatting and explain the notation. $\endgroup$
    – JAustin
    Commented May 29, 2016 at 15:53
  • 5
    $\begingroup$ And I believe it is oscillatory. The amplitude function is oscillatory, and the Mathematica plot is certainly periodic: imgur.com/YaLRuNq. $\endgroup$
    – JAustin
    Commented May 29, 2016 at 16:16
  • 1
    $\begingroup$ Do you mean $O(\theta^5)$? $\endgroup$
    – kennytm
    Commented May 30, 2016 at 3:31

The problem here isn't so much calculus as the assumptions that are made about the system. The solution can only be as accurate as the assumptions made, no matter how accurate the solutions of the equations are.

For a pendulum where the distance between the centre of gravity and the anchor point is $l$ and the mass is $m$, the equation of motion is:

$$\frac{d^2\theta}{dt^2}+\frac{g}{l}\sin \theta=0$$

Where $\theta$ is the angle between the pendulum and the vertical.

But this is mathematically difficult to solve, so we invoke the small angle approximation:

$$\sin \theta \approx \theta$$

This makes the equation easy to solve but its solutions only approximate.

Similarly for spring mass systems we usually assume the spring to be Hookean but for many real world systems that's only an approximation.

Often other assumptions about an oscillating system, such as no friction/no drag will also introduce further inaccuracy of the model vis-a-vis reality.


For a pendulum, you use the approximation $\sin(\theta)\approx\theta$ in the derivation of the simple harmonic equation of motion, which is only valid for small angles.

For a spring, it is Hooke's law itself that is only valid for reasonably small streching of the spring - a spring to which you apply hundred times the force needed to strech it by 10% usually will not be ten times its length, if it is still in one piece at all.


One has to keep clear the difference between the mathematical model proposed, and the physical system that it will describe.

As far as I know, SHM came from the differential equations of Hooke's law - so, using calculus, it should be really accurate.

The mathematical equation is accurate, because that is the function of mathematics.

The question is "can it model a specific physical setup?"

For a pendulum, you use the approximation sin(θ)≈θ in the derivation of the simple harmonic equation of motion, which is only valid for small angles.

The statement "is only valid for small angles" defines the bounds where the harmonic oscillator mathematical model can be used to describe the pendulum.

By the way, the mathematical harmonic oscillator in the quantization regime is very important because most symmetric physics potentials expanded in a taylor series have as a first dominant term the x^2 . That is why it is a useful potential for a lot of approximations of complicated potentials.


The differential equation for the harmonic oscillator $$ \frac{d^2x}{dt^2} - kx = 0 $$ is an approximation to the general pendulum. Going back to the elementary pendulum we have the basic sum of forces for the y coordinate oriented along the string and the x axis perpendicular to it. The forces are labelled in the diagram here

elementary pendulum

We do our sum of forces along the y direction, which is zero so the tension $T = mg\cos\theta$, and the acceleration in the $x$ direction is $$ ma = mg\sin\theta. $$ We usually make the following assumption. This angle is assumed to be small so we have $\sin\theta \simeq \theta$. The arc length of the swing is $r = \ell\theta$. for $\ell$ the length of the massless string. This differential equation, where $a = d^2x/dt^2$, is then written as $$ \frac{d^2\theta}{dt^2} - \frac{g}{\ell}\theta = 0, $$ which gives us the elementary solutions $\theta(t) = \exp\left(\sqrt{g/\ell}~t\right)$.

We return though to the general case with $$ \frac{d^2\theta}{dt^2} - \frac{g}{\ell}\sin\theta = 0, $$ This is the Jacobi equation, and its solutions are $$ \theta(t) = 2J_{\mathrm{am}}\left(\frac{1}{2}\sqrt{(k - 2g/\ell)(t + t_0)}\Big|\frac{4g/\ell}{2g/\ell - k}\right) $$ wnere $J_{\mathrm{am}}(x|y)$ is the Jacobi amplitude, or elliptic function. I am not going to go into that in any greater detail, but leave that to the reader.

The harmonic oscillator is a linearization of a more general problem. There are related subjects such as the Mathieu equation and Floquet solutions, and in a relativistic framework a related sin-equation is the sine-Gordon equation for a soliton.


All equations of motion are inaccurate to some extent. The notion of coordinate, time, force, etc. are inaccurate as a matter of fact and it is because of us. It is we who agree to use simplified, idealized notions for our purposes. For example, there are no two absolutely equal apples so we cannot count them as two apples, strictly speaking. But with accepting their differences as inessential to us, we start to apply math and count different things as equal. Math is our idealization and simplification for practical purposes.


These approximations are due to the fact that any potential can be approximated to the spring potential near its equilibrium points, which I shall prove.

Suppose you have a potential $V(x)$ and if $x_0$ is a stable equilibrium point in $V(x)$, then the potential near $x_0$ will be (Taylor Expansion), $$V(x_0 + dx) = V(x_0) + V^{'}(x_0)dx + \frac{1}{2}V^{"}(x_0)dx^2 + O(dx)$$ where $O(dx)$ denote higher order terms in $dx$. Since $x_0$ is an equilibrium point, $V^{'}(x_0)=0$ and since we are talking about points near $x_0$, $O(dx)=0$. Taking $V(x_0)=0$ to be the reference point, we have, $$V(x_0 + dx) = \frac{1}{2}V^{"}(x_0)dx^2$$ But this leads to the equation, $$md\ddot x = -V^{"}(x_0)dx$$ $$\Longrightarrow \omega = \sqrt{\frac{V^{"}(x_0)}{m}}$$ From this equation, you can derive angular frequency for all SHMs. But notice that we had neglected the higher order terms of $dx$ since we assumed that $dx$ is very small. This assumption will be present in some form or the other in all the different derivations of the formula for SHMs. For the spring, it will be in the Taylor expansion of $\sin(\theta)$, for springs it is the approximation in Hook's law. It is due to this approximation, that the formulas for most SHMs are not valid for large amplitudes.


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