Explore the Link between Pendulums and Springs

[sponsoredwalk]

Bear with me if you've read or done a problem like this before, I just thought of it while
watching the lecture on Hooke's Law http://ocw.mit.edu/courses/physics/8-01-physics-i-classical-mechanics-fall-1999/video-lectures/lecture-10/ by Walter Lewin of M.I.T. and
probably couldn't solve it but I thought I'd ask what would happen anyway.
If it's too boring just skip to the -------------- part below and if you need to check back
I wrote it out just in case so have a look at it.

$$F \ = \ ma \rightarrow \ - \ k \ x \ = \ m \ \ddot{x}$$

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

You search and search and discover $$x \ = \ A \cos( \ \omega \ t \ + \ \phi \ )$$
will be a solution leading you to discover $$\omega \ = \ \sqrt{ \frac{k}{m} }$$

If you were dealing with a situation like that in this picture:

[PLAIN]http://img830.imageshack.us/img830/1043/springx.jpg

You'd set it up like so:

$$F_x \ = \ ma_x \rightarrow \ - \ T( \theta) \ \sin \ ( \theta ) \ = \ m \ \ddot{x}$$

$$F_y \ = \ ma_y \rightarrow \ T( \theta) \ \cos \ ( \theta ) \ - \ m \ g \ = \ m \ \ddot{x}$$

We can then reduce $$F_x$$ to $$- \ T( \theta ) \ \frac{x}{l} \ = \ m \ \ddot{x}$$

Using small angle approximations we can see that the y-direction acceleration will be
negligible and cosθ will be close to 1. This let's us set T = mg and solve the other equation.

$$- \ m \ g \ \frac{x}{l} \ = \ m \ \ddot{x}$$

Leading to:

$$\ddot{x} \ + \ \frac{g}{l} \ x \ = \ 0$$

From this the angular frequency is shown to be $$\omega \ = \ \sqrt{ \frac{g}{l}}$$
The time period is calculated to be $$T \ = \ \frac{2 \ \pi}{ \omega} \ = \ 2 \ \pi \ \sqrt{ \frac{l}{g} }$$

This shows us that the time period of a pendulum is really determined by the length of
the pendulum when dealing with a small angle.

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My question is if we set up a situation like that in the picture but instead of a rigid
arm or string with a mass on the end we used an ideal spring with a mass on the end
and let it follow the same path as a pendulum only the spring itself was launched into
Simple harmonic motion. Would the situation be the same?

I'm thinking a good question would be, how long does the pendulum arm connected
to the spring have to be in order for both the pendulum and the spring to oscillate
in unison, i.e. when the pendulum reaches the lowest point the spring is at it's
closest to the pendulum and when the pendulum is fully to the right/left the spring
is stretched furthest. The green line with the "x" in the second picture could also
act as the equilibrium line for the spring, just for a reference point when visualizing this.
I should have drawn it wider, the spring is actually compressed when the pendulum
is vertical.

[PLAIN]http://img547.imageshack.us/img547/1663/crazy.jpg

The spring is itself massless so can I use a similar analysis to that of the above example?
Will the mass + spring moving up and down alter the $$T \ = \ 2 \ \pi \ \sqrt{ \frac{l}{g} }$$ period of the entire system?
If not I could use $$K.E. \ + \ U_{grav} \ + \ U_{spring}$$ on the mass/spring
and figure out how long the length L of the pendulum would have to be to get this
working right?

You see I'm just trying to synthesize this so I apologise if it's a bit shoddy!
I'm probably not accounting for something, like the force you apply to stretch
the string in the first place messing the situation up. If that is the case you could
balance it by applying an equal force where the spring is connected to the
pendulum to balance it though? No?

If I've messed this setup up then correcting it to work rather than salvaging my bad
attempt would be a lot better ! Thanks for reading it.

 

[rock.freak667]

I've probably done this before in vibrations class a long time ago, but you can use the formula:

$$\frac{dU}{dt}+\frac{dT}{dt} = 0$$

Where U is the PE of the system and T is the KE of the system

and get the equation of motion for the system.