# Why is Simple Pendulum not SHM?

Why does Simple pendulum's motion not hold as SHM for large angles, only for small angle approximations? The restoring force is still directed towards the mean position. I know mathematically the reason is because at large angles it is not linearly proportional to displacement, but how do I make sense of it intuitively?

• Are you asking why the term simple harmonic motion applies only when $F = -kx$ and not to systems that deviate slightly from this? Commented Jan 23 at 13:54
• And because $x \neq \sin x$ Commented Jan 23 at 13:55
• Solution to pendulum differential equation Commented Jan 23 at 16:01

I would say there's not an intuitive reason for it because simple harmonic motion is a mathematical concept, not an intuitive one.

The big intuitive concept is "oscillatory motion" which is what results when an object feels a restoring force towards a point. Pendulums, springs, and orbiting planets all fit into this category.

"Simple harmonic motion" is an extremely specific kind of oscillatory motion that only happens when the strenght of the restoring force is directly proportional to the distance from the center point. That's its definition. As Frederic Thomas' answer points out, this basically never happens in the real world, though it's a very useful approximation for many systems.

While there are good reasons to expect that many systems will be approximate simple harmonic oscillators, there is no good reason to expect that any system will be exactly a harmonic oscillator because it has such a specific mathematical definition.

Here, we give a nice physical reason, in terms of forces and accelerations, for why and how a pendulum is different than a simple harmonic oscillator. The tldr is that the restoring force increases linearly with amplitude for a SHO, but for a pendulum, the force actually decreases down to zero once the amplitude gets large enough. This results in the pendulum spending much more time at the full amplitude positions than an oscillator when the amplitude is large enough.

We know that simple harmonic motion, described by the differential equation $$\ddot{x} = -\omega^2 x$$ is solved by sinusoidal functions, $$x(t) = A\sin(\omega t+\varphi)$$. Notably, at the turning points (positions where the oscillator is at its maximum displacement from equilibrium and hence instantaneously stopped), the force is the largest, meaning that the curvature of the function is maximal, and this is true no matter how large the amplitude is. This gives rise to a function that is more sharply curved at minima and maxima, and this graphical fact represents the fact that it doesn't take a long time for the oscillator to come to a stop, turn around, and move back in the other direction.

Now, imagine a pendulum with a thin, rigid rod in place of a string. The restoring force on the mass at the end of the rod is the component of the gravitational force pointed tangent to the circle in which the mass moves. This restoring force increases approximately linearly as the amplitude grows from zero. However, once the angle gets large enough, this force no longer grows linearly. In fact, the restoring force becomes a maximum when the rod is held horizontal to the ground, when the amplitude is $$\theta_0 = \pi/2$$ (90 degrees relative to the vertical).

Then, as we continue to increase the amplitude above $$\theta_0=90$$ degrees, the restoring force shrinks as the component of the gravitational force tangent to the circle decreases. When the rod is very nearly vertical, the force is very nearly zero! This is markedly different than the harmonic oscillator case, in which the restoring force continues to grow as the amplitude grows.

Finally, since the force is very small when the rod is nearly vertical, the acceleration is very small, and so it takes a long time for the object to slow down, come to a stop, turn around, and start moving in the other direction. This matches the graphical fact that the second derivative is small (because the acceleration is the second derivative!) and hence the curvature of the graph there is small, meaning that the graph is sort of flat.

We can see all of this in the picture below, in which I've plotted (in red) the motion of a rigid pendulum with an initial amplitude of $$\theta_0 = \pi - 0.01$$ (very nearly vertical!) as a function of time. I've also included the corresponding sinusoidal function (black, dashed) for reference. We can see that the graph is very flat (relative to the sinusoidal one) at the maxima and minima, indicating that the pendulum spends a long time at the top of the motion, exactly because the restoring force there is very small.

The curved surface in a small region appears flat, similar to how Earth, despite being a deformed sphere, can be approximated as a flat surface on a playground. Likewise, an arc of a circle can be approximated as a straight line for a small angle. This is why the motion of the pendulum bob can be approximated as linear motion for small angles. However, for large angles, the linear approximation is not applicable. In simple harmonic motion (SHM), the restoring force is proportional to the displacement from the equilibrium position. While one can consider the motion of a simple pendulum for large angles, the nonlinear terms of displacement need to be taken into account.

As the OP knows the mathematical reason for a pendulum being a non-perfect oscillator, I will stay in the explanation very general.

The Simple Harmonic Oscillator is an idealization. I would say, the perfect harmonic oscillator practically does not exist. Similar to motion without friction. On earth it does not exist, may be in open space in the universe.

The oscillator always comes along with small nonlinearities, as for instance in the pendulum. Nonlinearities in particular appear at larger amplitudes since certain assumptions on a particular dependency are no longer valid. A general function can be developed in a Taylor series:

$$f(x) = a + b\cdot x + c\cdot x^2 + \ldots$$

If $$x$$ is small, i.e. $$x<1$$ we can neglect the higher order terms $$x^2\ldots$$. We can then calculate with $$f(x) = a +bx$$. However, if $$x>1$$ we cannot longer neglect the term $$c\cdot x^2 + \ldots$$, it will have essential influence on the function's behaviour.

Nonlinear oscillations are more interesting than purely linear ones. Actually, our life is determined by nonlinear effects (would we even exist if the world would be only linear BTW), and physics is often not able to describe them. If air behaved only according linear laws, we could make weather forecast for years. But we can't do that since the laws which describe the motion of air are highly nonlinear. Actually this happens all the time and everywhere. The SHO is a mathematical construct, but not reality. Reality is different and more or less nonlinear.