# Pendulum and spring physics

1. Mar 31, 2005

### T@P

I read about this on another post, and it got me interested. suppose you have a pendulum (say a steal bar with a weight at the bottom or something) and a spring attached to the bottom. now you take the spring and pull it all the way down. you should have the pendulum pointing straight down, and the spring extended unerneath.

what happens when you release the spring? note: it is not an 'ideal' pendulum, i.e. it is not pointing 'straight' down. therefore the spring doesnt just oscilate up and down, the pendulum moves a bit too i think.

2. Mar 31, 2005

### hemmul

what is the mass of the spring in comparison with the pendulum?

3. Mar 31, 2005

### Integral

Staff Emeritus
This is an example of a coupled system, how it behaves will depend upon the length of the pendulum, the mass on the spring and spring constant. What you will see is an exchange of energy between oscillations of the pendulum and the spring system. Thus the motion will be complex, at some point in time it will be pure pendulum motion at another point in time it will be simple spring oscillations, at other times some combination of the 2.

4. Mar 31, 2005

### jdavel

Integral,

This isn't really a "coupled system"; there's only one mass. The force (magnitude and direction) on the mass is uniquely determined by its location.

5. Mar 31, 2005

### T@P

well is there a general formula for what happens? and also, nothing especially wacky would happen? (too bad)

6. Mar 31, 2005

### Integral

Staff Emeritus
I am assuming that there is a mass at the end of the pendulum to which a spring and mass is attached to. Even if there were no mass at the end of the pendulum rod there would still be some coupling cross the densitiy change at the rod/spring interface.

7. Mar 31, 2005

### Crosson

Yes there is a formula which is very simple to apply, but it has to do with differential equations. If you have had differential equations, then you should have no problem understanding me when I say:

Define the Lagrangian L:

L = kinetic energy - potential energy

So your Lagrangian will be a function of the variables:

$$L = L (x, v , \theta , \omega )$$

Which are, in order: Position of the spring, velocity of the spring, (angular) position of the pendulum, (angular) speed of the pendulum.

These are the four variables that you would like to know as a function of time. To find the functions, you need a (differential) equation of motion. Here it is:

$$\frac{\partial{L}}{\partial x} - \frac{d}{dt} (\frac{\partial{L}}{\partial v}) = 0$$

$$\frac{\partial{L}}{\partial \theta} - \frac{d}{dt} (\frac{\partial{L}}{\partial \omega}) = 0$$

So, just plug and chug :)

8. Mar 31, 2005

### Integral

Staff Emeritus
Actually the motion should be very interesting, and since the system would/could be non linear you may find chaotic.

This type of system is often solved using Lagrangian Dyanmics. See this thread for an example of how this works.