1
$\begingroup$

If we use a helical spring instead of string in a simple pendulum then will the body execute two simple harmonic motions simultaneously? Like up-down motion of spring and to and fro motion of simple pendulum.

$\endgroup$
  • $\begingroup$ en.wikipedia.org/wiki/Lissajous_curve $\endgroup$ – mmesser314 Mar 22 '15 at 15:25
  • $\begingroup$ It would be pretty interesting to make one... I'll be back... $\endgroup$ – Jimmy360 Mar 22 '15 at 16:32
  • $\begingroup$ Well @Jimmy360 I have made. It works like a charm $\endgroup$ – Saad Mar 23 '15 at 0:43
0
$\begingroup$

In general

Yes! A simpler example might be a mass in a 2-D plane attached to four springs (one on each side). Assuming the oscillations are small, the motion of the mass can be described as the combination of oscillations in both $x$ and $y$ directions.

Solving your example

In fact, in your specific example, we do see that phenomenon (for small oscillations!).

Assume we have a pendulum where the string is actually a spring. For simplicity, I'll restrict the motion of the pendulum to one plane (ie it can swing in one direction, but is blocked along the other).

The total energy can be written as the sum of kinetic energy (in both the swing direction and radial direction), and the gravitational potential energy and the spring potential energy.

$E(l, \theta)=\frac{1}{2}m\dot{l}^2+\frac{1}{2}ml^2\dot{\theta}^2+\frac{1}{2}k(l-l_0)^2+mg(l_0-l\cos\theta)$

where $l$ is the length of the pendulum spring, $l_0$ the "natural" length of the pendulum spring if you lay it flat on a table, $\theta$ is the angle from vertical, dots indicate time derivatives. I set the $E=0$ reference for the gravitational part as when the pendulum is at the bottom of its motion.

Now let's assume the swinging motion is small: if we Taylor expand the $\cos\theta$, keeping only up to second-order in small $\theta$:

$E(l, \theta)=\frac{1}{2}m\dot{l}^2+\frac{1}{2}ml^2\dot{\theta}^2+\frac{1}{2}k(l-l_0)^2+mg(l_0-l)+\frac{1}{2}mgl\theta^2$

By completing the square, we can combine the third and fourth term, and (ignoring a constant term that appears and just shifts the energy), we find

$E(l, \theta)=\frac{1}{2}m\dot{l}^2+\frac{1}{2}ml^2\dot{\theta}^2+\frac{1}{2}k\left(l-l_0-\frac{mg}{k}\right)^2+\frac{1}{2}mgl\theta^2$

Let $x=l-x_0$ where $x_0=l_0+\frac{mg}{k}$

$E(x, \theta)=\frac{1}{2}m\dot{x}^2+\frac{1}{2}kx^2+\frac{1}{2}m(x+x_0)^2\dot{\theta}^2+\frac{1}{2}mg(x+x_0)\theta^2$

The first two terms look like our usual harmonic oscillator in $x$, oscillating around $x=0$.

We note that, if the swing is small ($\theta$ is small) and the spring movement is small ($x$ is small compared to $x_0$), then $x\dot{\theta}^2$ and $x\theta^2$ have three factors of smallness, and $x^2\dot{\theta}^2$ has four! I'll get rid of the $x^2\dot{\theta}^2$ term since we already killed off some terms with four factors of smallness when expanding $\cos\theta$. Now, rearranging:

$E(x, \theta)=\frac{1}{2}m\dot{x}^2+\frac{1}{2}kx^2+\frac{1}{2}mx_0^2\dot{\theta}^2+\frac{1}{2}mgx_0\theta^2 + P(x, \theta)$

where $P(x, \theta)=mxx_0\dot{\theta}^2+\frac{1}{2}mgx\theta^2$.

Interpreting

Forget about $P$ for a moment, and you see we have two independent harmonic oscillator motions happening at once:

  1. The spring length oscillates with frequency $\omega_s=\sqrt{k/m}$ about the point $l=l_0+mg/k$
  2. The pendulum swings with frequency $\omega_p=\sqrt{\frac{g}{l_0+mg/k}}$ around $\theta=0$

So, to first order, it is roughly what you would guess, two harmonic motions, but for a pendulum/spring of natural length $l_0+mg/k$ (ie the spring is stretched by gravity).

Including perturbations

Let's see what effect the first-order perturbations ($P$) might have on this motion in a couple of easy limits. This is really outside the bounds of your question, and a bit technical, but it's the sort of thing you might find interesting later on in your physics learning. And I've always found that looking ahead is part of what makes physics so exciting. So don't worry if it doesn't all make perfect sense yet.

First, let's assume the pendulum mode is really fast compared to the spring mode, and just estimate how the pendulum mode affects the spring mode by ``averaging the pendulum mode out of our problem'' to write an energy $E(x)$ just in terms of $x$. We are basically saying that at each spring position $x$, the pendulum is making many many swings, much faster than the spring is moving to the next position, so all the spring motion cares about is the time-average of the pendulum energy.

The time-average of the $\frac{1}{2}mx_0^2\dot{\theta}^2+\frac{1}{2}mgx_0\theta^2$ part doesn't depend on $x$, so it's just an energy shift as far as we're concerned. But the perturbation terms do have $x$ in them. If $\theta$ is duly executing simple harmonic motion with frequency $\omega_p$, you can show with some trigonometry that $P(x)=\frac{3}{2}mg\langle\theta^2\rangle x$, where $\langle\theta^2\rangle$ is the time-averaged square angle. So, if we play the completing the square game again with $x$, then we find

$$E(x)=\frac{1}{2}m\dot{x}^2+\frac{1}{2}k\left(x+\frac{3mg}{2k}\langle\theta^2\rangle\right)^2$$

That is to say, we have a harmonic oscillation around $l=l_0+\frac{mg}{k}-\frac{3mg}{2k}\langle\theta^2\rangle$. Interpretation: having a little oscillation into the pendulum mode makes the spring want to tighten in a little bit, because the tighter the spring goes, the less energy the pendulum oscillation contributes.

(I should note that, in the opposite limit, where the spring mode is really fast compared to the pendulum mode, it doesn't have any first order effect on the pendulum mode, because the time-average over $x$ oscillating is zero.)

What happens dynamically

But looking at the average effect one mode has on another misses out on the dynamic effects. That is, if you put energy into one mode (say, start the pendulum swinging), some of it might sneak over to the other mode, and sneak back and forth.

How can we see this? Think of our first limit, $\omega_p\gg \omega_s$. Imagine the mass is at rest at $\theta=0, l=x_0$. Then you poke the mass so it starts swinging a little. Now, from above, the spring will want to tighten a bit to reduce that oscillation energy, so the spring mode will move a bit, but that starts it oscillating as well. Now both modes are playing together, exchanging energy.

(And I should note that, in the opposite limit, even though we found that there was no first order change in the pendulum motion due to the spring motion, the spring mode can still feed the pendulum mode some energy if it is already started moving. This phenomenon is called parametric oscillation.)

Anyhow, I hope you enjoyed my tangent, because I had some fun thinking about it. And I apologize if there were any typos in the math!

$\endgroup$
  • $\begingroup$ Cool!! I didn't get what u r trying to say coz i'm just a high school student. But anyways cool $\endgroup$ – Saad Mar 23 '15 at 0:46
  • $\begingroup$ It can be with just two springs.One for each independent dimension(z and y) although it is the same thing because you put two springs in each direction with half K each.But it is unnecessary. $\endgroup$ – TheQuantumMan Mar 23 '15 at 0:57
  • $\begingroup$ True, I just find it easier to visualize with four springs since it makes the problem more visually symmetric (and leads more naturally into thinking about phonons). But two springs is fine! $\endgroup$ – Sam Bader Mar 23 '15 at 1:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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