# Is the period of a harmonic oscillator really independent of amplitude?

Say we have a harmonic oscillator that obeys the force rule: $$F=-kx$$ Hence, the equation of motion is: $$\ddot{x}+\frac{k}{m}x=0$$ which may be solved analytically as: $$x(t)=x(0)\cos\left(\sqrt{\frac{k}{m}}t\right)+\frac{\dot{x}(0)}{\sqrt{\frac{k}{m}}}\sin\left(\sqrt{\frac{k}{m}}t\right)$$ from which it is clear that the period of its oscillation in time is given by $$T=2\pi\sqrt{\frac{m}{k}}$$. Now, as I understand the term "amplitude", it refers to the maximum displacement of the harmonic oscillator from its equilibrium position, which in this case is simply the origin of the $$x$$-axis. It follows that by setting $$\dot{x}(t):=0$$, finding the value of $$t$$ that satisfies this (there are infinitely many, but we just need one), and then plugging that value of $$t$$ into the function $$x(t)$$, I should get the amplitude of the oscillation. If we call that special $$t$$-value $$t=t^*$$, here is what I get: $$x(t^*)=\frac{x(0)|x(0)|\sqrt{\frac{k}{m}}+\dot{x}(0)|\dot{x}(0)|\sqrt\frac{m}{k}}{\sqrt{\dot{x}(0)^2+x(0)^2\frac{k}{m}}}=A$$ where $$A$$ stands for amplitude. But if you make the substitution $$T=2\pi\sqrt{\frac{m}{k}}$$ everywhere you can in that expression, then you can check that $$\frac{\partial{A}}{\partial{T}}\neq{0}$$, and therefore it seems that the statement "period is independent of amplitude" (also known as the property of isochronism) is wrong? One way I found to make that statement correct is to assume that the harmonic oscillator starts at rest (i.e. $$\dot{x}(0)=0$$), but I'm not sure if people only meant for that phrase to apply to this starting-at-rest situation or if it was intended to have a more general applicability.

• You must have a mistake somewhere. Consider that it's always at rest at some point on the path, and you could have started your stop-watch at that point. So you should absolutely not find that the answer depends on the value of $\dot{x}(0)$. Commented Mar 30, 2021 at 14:59
• Your are misreading the broad statement, which indicates you may vary period and amplitude independently in a solution. It is not that amplitude and period are independent for the very same solution! Commented Mar 30, 2021 at 17:32

The period is independent of the amplitude keeping $$k$$ and $$m$$ fixed (and varying initial conditions). Your calculation tries to keep the initial conditions fixed and vary $$k$$ and $$m.$$

• Can you elaborate your answer by showing where I might have been varying $k$ and $m$? I certainly don't have the impression that I was doing that, since when I differentiated $x(t)$ with respect to $t$, I treated $\sqrt{k/m}$ as a constant...(if that's what you meant). Commented Mar 30, 2021 at 16:09
• @SurfaceIntegral You are trying to find $A$ as a function of $x(0)$ and $\dot{x}(0)$. To go from your initial position and velocity to the amplitude requires knowing the spring constant and the mass. So for fixed $x(0)$ and $\dot{x}(0)$, $A$ will depend on $k/m$ (equivalently, $A$ will depend on $T$). Essentially, when you take $dA/dT$ while assuming $T$ is proportional to $\sqrt{m/k}$, you are necessarily varying $k$ and $m$ while keeping $x(0)$ and $\dot{x}(0)$ fixed. Commented Mar 30, 2021 at 16:54
• @surface Later on, you take a derivative with respect to $T,$ a function of $m$ and $k.$ This is not meaningful if $m$ and $k$ are constant. Commented Mar 30, 2021 at 17:28

I'm not sure why you're making this so complicated. Write \begin{align} x(t)=A\cos(\omega t)+B\sin(\omega t) \, ,\tag{1} \end{align} where $$\omega^2=k/m$$, $$A=x(0)$$ and $$B= \dot{x}(0)/\omega$$. Since $$x(0)$$ and $$\dot{x}(0)$$ are arbitrary, so are $$A$$ and $$B$$.

Now rewrite (1) as \begin{align} x(t)=C\cos(\omega t-\varphi)=C\cos(\varphi)\cos(\omega t)+C\sin(\varphi)\sin(\omega t) \end{align} so clearly $$A=C\cos(\varphi)$$ and $$B=C\sin(\varphi)$$.

Obviously in this form the amplitude $$C=\sqrt{A^2+B^2}$$ is arbitrary (since $$A$$ and $$B$$ are arbitrary) and does not depend on $$\omega$$. Now the amplitude $$C$$ depends on the initial conditions through \begin{align} C=\sqrt{x^2(0)+\dot{x}^2(0)/\omega^2} \tag{2} \end{align} but you still have the freedom to choose $$x(0)$$ and $$\dot{x}(0)$$ to make $$C$$ whatever you want, irrespective of $$\omega$$.

• Neat avatar. I added an obvious deconstructive comment to the OPs question, but I could move it here, if you wished; or, better, you could adopt it as your punchline. Commented Mar 30, 2021 at 19:53