# What are the possible limits for when SHM is not applicable for a vertical mass-spring system?

I am not sure if this is more of a maths problem than a physics if so could admin place in the math stack.

So my question is as follows. I have recently been looking at SHM in a spring-mass, as shown by the picture and it got me thinking about the equation of motion of SHM the conditions needed for SHM to occur specifically the maximum and minimum value of displacement.

What I mean but maximum and minimum values are like so. A condition for SHM is that the acceleration of the object is proportional to displacement, and is shown as:

$$a_x\propto x$$

But having a little think, there must be a minimum displacement and maximum displacement for this to occur.

But using N2L and finding the net force of the system when the mass has been displaced we get the following. So if I displace the mass by a very small displacement, that would not necessarily mean that we would get SHM, and vice versa if I displace by a very large amount this would not necessarily give SHM, so in my mind, there must limit on the amount you can displace the object by.

I have however been digging through textbooks and nothing really address this which make me think, that it either plane obvious and I am missing a fundamental assumption, or that there is no reason for it to be addressed.

Anyhow I did try and think of a way I could explore this, and hit a bump.

If I take the equation of motion for SHM and replace the 'x' in the k x portion of with ($$x_0+\delta x$$)

$$\ddot x+\frac{k}{m}x=0$$

$$\ddot x+\frac{k}{m}(x_0+\delta x)=0$$

now I thought maybe I could binomially expand this as it would give me some insight into the maximum and minimum values but now I don't' think so much as I can see how the expansion is giving me any insight.

$$\left(x+\delta x\right)\approx x\left(1+\frac{\delta x}{x}-\frac{\delta \:x^2}{x^2}+2\frac{\delta \:x^3}{x^3}...\right)$$gy

tl;dr

Does anyone have any insight on how we can show mathematically where the >limits are for the relationship for $$a \propto x$$ to break down.

• If you deform the spring, then you're no longer able to use $F=-kx$, does that count as a limit? – Kyle Kanos Dec 21 '18 at 13:49
• yes but, what about a small displacement? – james2018 Dec 21 '18 at 13:54
• Other than the displacement being zero, I don't know why SHM wouldn't occur for arbitrarily small displacement. You might need to explain why you think it wouldn't happen. – Kyle Kanos Dec 21 '18 at 13:56

Although I am quite confused as to where exactly your problem in understanding is, the limits that you're talking about do indeed exist in oscillatory motion. But the example you chose - a spring following Hooke's Law, the SHM differential equation is always valid, for all $$x$$. However, as one might intuitively expect, pulling a spring too hard will evoke a very different response than S.H.M type oscillation. In other words, $$F = -kx$$ is no longer the force law to describe such a deformed-spring system. Another example to show you how such 'limits' of S.H.M are seen physically, imagine a simple pendulum. You can easily show that it's motion follows the differential equation $$\ddot\theta + \frac{g}{l}\sin\theta = 0$$, where $$\theta$$ is the angle made by the pendulum with the vertical. This is a non-linear equation. Only for small $$\theta$$ can we approximate the differential equation to the familiar S.H.M equation with nice-looking solutions. For large $$\theta$$ such an approximation does not describe the physics of the pendulum accurately. The limits that you wish to find does not arise from the S.H.M equation itself, but from more complicated differential equations, which behave linearly under some assumptions.

• "a spring following Hooke's Law, the SHM differential equation is always valid" - that is only true for a massless spring. Real springs have mass, inertia and store kinetic energy as well as potential so the SHM equation is not valid. – ja72 Dec 21 '18 at 16:31
• @ja72 Well, for idealized, perfectly-Hookean, massive spring you can simple add $m_\text{spring}/3$ to the nominal mass of the oscillator and get back to SHM. – dmckee --- ex-moderator kitten Dec 22 '18 at 0:34
• @dmckee - the $\frac{1}{3}$ factor only applies for near zero excitation frequencies where $\ell \omega / c \ll 1$ (here $c$ is speed of sound in medium). Just for fun, try to derive this rule and you will find many more details and twists than you expect. – ja72 Dec 22 '18 at 16:14

Experimenter's answer: your vertical SHM lab will not yield expected results (even accounting for the spring mass) if ...

• You stretch the spring beyond the elastic limit. This disaster is easily detected, because the spring will by misshapen after the experiment. Get a new spring and use a smaller amplitude.

• You are using loose weights on a hanger and the downward acceleration ever exceeds $$g$$ You can detect this event because you will hear the plates clinking together partway down each oscillation. Use a smaller amplitude, more mass, or a lighter spring.

• The spring coils touch during the compressive motion This generally induces the afore-mentioned negative acceleration and clinking.

In general it tends to be easier to think in terms of potential than forces for those problems. The force is then the negative derivative of the potential.

If you have a potential $$V(x)$$, first find the equilibrium position (or positions), which correspond to local extrema of the potential where $$dV/dx=0$$. Suppose $$x_0$$ is a local minimum. then expand the potential about this minimum \begin{align} V(x_0+\delta x)&\approx V(x_0)+\delta x \frac{dV}{dx}\vert_{x=x_0}+ \frac{1}{2}\delta x^2 \frac{d^2V}{dx^2}\vert_{x=x_0}\, ,\\ &=V(x_0)+ \frac{1}{2}\delta x^2 \frac{d^2V}{dx^2}\vert_{x=x_0} \tag{1} \end{align} since by construction $$dV/dx=0$$ at $$x=x_0$$. The condition for a minimum is that $$\frac{d^2V}{dx^2}\vert_{x=x_0}>0$$. Indeed, (1) is an osculating parabola going through $$x_0$$ for the potential $$V(x)$$.

Now, $$F=ma$$ takes the form $$m\delta\ddot{x}=-\left(\frac{d^2 V}{dx^2}\vert_{x=x_0}\right)\delta x$$ with solution $$x(t)=x_0+\delta x(t)=x_0+A\cos(\omega t +\phi)$$ with $$\omega^2=\frac{1}{m}\left(\frac{d^2 V}{dx^2}\vert_{x=x_0}\right)$$.

Note that, if $$\left(\frac{d^2 V}{dx^2}\vert_{x=x_0}\right)<0$$, the equilibrium position is unstable (there is a local max in the potential) and there is no harmonic motion since the solutions are exponentials near $$x_0$$: basically the force pushes the particle away from the equilibrium position.

In your specific case, the potential is, up to a constant, $$\frac{1}{2}kx^2+mg x$$, where $$x$$ is measured from the unstretched position of the spring. The effect of gravity is to displace the equilibrium position without affecting the frequency of oscillation. Since your potential is always quadratic, there is no limit to the motion although physically the spring will start to display non-linearity for large values of $$x$$, i.e. you cannot compress it ad infinitum else it would crash back into its holder. The range of displacements about which the spring remains mostly linear depends on the actual spring or the properties of the device that supplies the restoring force (v.g. a plastic ruler slightly bent will produce a linear restoring force, but typically over a rather small range.)

The strategy generally works for any potential, including molecular potentials such as Morse, which is of the type $$A (e^{-2\alpha x}-2 e^{-\alpha x})$$ (where $$A$$ and $$\alpha$$ are parameters which depend on the molecule). This potential does have harmonic oscillations near the bottom of the well. Keeping more terms in the expansion of the potential leads to anharmonic effects, i.e. the result is not quite harmonic motion.