[Physics] the time period of an oscillator with varying spring constant

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It is well known that the time period of a harmonic oscillator when mass $m$ and spring constant $k$ are constant is $T=2\pi\sqrt{m/k}$.

However, I would be interested to know what the time period is if $k$ is not constant. I have searched hours after hours for right answers from Google and came up with nothing. I am looking for an analytical solution.

Best Answer

Here is a solution for a spring force that varies directly with displacement. It thus varies with time implicitly, but has no explicit dependence on time or any other variable.

Givens and Assumptions

  • oscillator with mass $m$
  • amplitude of oscillation $A$
  • oscillator displacement, $x$, varies with time, but $x(t)$ is unknown
  • spring applies force varying with displacement, $F(x)$
  • The function $F(x)$ is an odd function, that is $F(-x) = -F(x)$ (otherwise the amplitude could be different in the positive and negative directions - see below for what to do in this case)
  • equilibrium position is $x=0$, that is $F(0) = 0$ (for convenience only)

Objective

Find the period of oscillation, $T$

Solution

Starting from conservation of energy, the sum of the kinetic and potential energy of the mass must be equal to the total energy, which is constant. $$KE(x)+PE(x)=E$$ $$KE(x)=\frac{1}{2}mv^2(x)$$ $$PE(x)=\intop_0^xdx'\,F(x')$$ So $PE(x)$ is the potential energy stored in the spring, with $x'$ as just an integration variable.
We can think of $PE(x)$ as another way of defining the force-displacement relationship of the spring. We can define the potential energy versus displacement or the force versus displacement, and getting the other one is fairly easy.

Now, at $x=A$, $KE(x=A)=0$, so $PE(A)=E$ is known.

And so we have $$\frac{1}{2}mv^2(x)=PE(A)-PE(x)$$ Solving for $v(x)$, $$v(x)=\sqrt{\frac{2}{m}\left(PE(A)-PE(x)\right)}$$

Because $v=\frac{dx}{dt}$, we can also write $$dt=\frac{dx}{v(x)}$$ Integrating both sides, the time to go from a position $x_0$ to $x_1$ is $$\Delta t = \intop_{x_0}^{x_1}\frac{dx}{v(x)}$$ In particular, we know the time required to go from $x=0$ to $x=A$ is $T/4$, so $$T=4\intop_0^A\frac{dx}{v(x)}$$ $$T=4\intop_0^A\frac{dx}{\sqrt{\frac{2}{m}}\sqrt{PE(A)-PE(x)}}$$ which further simplifies to...

Final Result

$$T=\sqrt{8m}\intop_0^A\frac{dx}{\sqrt{PE(A)-PE(x)}}$$

Check of Result

For the linear case, $F(x)=kx$, so $PE(x)=\frac{1}{2}kx^2$ and $PE(A)=\frac{1}{2}kA^2$, which gives $$T=\sqrt{8m}\intop_0^A\frac{dx}{\sqrt{\frac{k}{2}}\sqrt{A^2-x^2}} =4\sqrt{\frac{m}{k}}\intop_0^A\frac{dx}{\sqrt{A^2-x^2}}$$ This integral can be looked up in a table, to obtain $$T=4\sqrt{\frac{m}{k}}\left(\sin^{-1}(1)-sin^{-1}(0)\right)=4\sqrt{\frac{m}{k}}\frac{\pi}{2}$$ $$T=2\pi\sqrt{\frac{m}{k}}$$ as expected. (QED)

We can dispense with the assumption that $F(x)$ is odd if we define two amplitude values: $A_+ > 0$ for the amplitude in the positive direction and $A_- < 0$ for the amplitude in the negative direction.

The total oscillator energy, $E = PE(A_+) = PE(A_-)$, so we can still call it $PE(A)$ as long as we remember what that means now.

Then, the period is twice the time required to go from $A_-$ to $A_+$, and so $$T=2\intop_{A_-}^{A_+}\frac{dx}{\sqrt{\frac{2}{m}}\sqrt{PE(A)-PE(x)}}$$ $$T=\sqrt{2m}\intop_{A_-}^{A_+}\frac{dx}{\sqrt{PE(A)-PE(x)}}$$

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