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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.

[tex] F \ = \ ma \rightarrow \ - \ k \ x \ = \ m \ \ddot{x} [/tex]

[tex] \ \frac{k}{m} \ x \ + \ \ \ddot{x} \ = \ 0 [/tex]

You search and search and discover [tex] x \ = \ A \cos( \ \omega \ t \ + \ \phi \ )[/tex]

will be a solution leading you to discover [tex] \omega \ = \ \sqrt{ \frac{k}{m} } [/tex]

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:

[tex] F_x \ = \ ma_x \rightarrow \ - \ T( \theta) \ \sin \ ( \theta ) \ = \ m \ \ddot{x} [/tex]

[tex] F_y \ = \ ma_y \rightarrow \ T( \theta) \ \cos \ ( \theta ) \ - \ m \ g \ = \ m \ \ddot{x} [/tex]

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

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.

[tex] - \ m \ g \ \frac{x}{l} \ = \ m \ \ddot{x} [/tex]

Leading to:

[tex] \ddot{x} \ + \ \frac{g}{l} \ x \ = \ 0 [/tex]

From this the angular frequency is shown to be [tex] \omega \ = \ \sqrt{ \frac{g}{l}}

[/tex]

The time period is calculated to be [tex] T \ = \ \frac{2 \ \pi}{ \omega} \ = \ 2 \ \pi \ \sqrt{ \frac{l}{g} } [/tex]

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.

--------------------------------------------------

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 [tex] T \ = \ 2 \ \pi \ \sqrt{ \frac{l}{g} } [/tex] period of the entire system?

If not I could use [tex] K.E. \ + \ U_{grav} \ + \ U_{spring} [/tex] 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.

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.

[tex] F \ = \ ma \rightarrow \ - \ k \ x \ = \ m \ \ddot{x} [/tex]

[tex] \ \frac{k}{m} \ x \ + \ \ \ddot{x} \ = \ 0 [/tex]

You search and search and discover [tex] x \ = \ A \cos( \ \omega \ t \ + \ \phi \ )[/tex]

will be a solution leading you to discover [tex] \omega \ = \ \sqrt{ \frac{k}{m} } [/tex]

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:

[tex] F_x \ = \ ma_x \rightarrow \ - \ T( \theta) \ \sin \ ( \theta ) \ = \ m \ \ddot{x} [/tex]

[tex] F_y \ = \ ma_y \rightarrow \ T( \theta) \ \cos \ ( \theta ) \ - \ m \ g \ = \ m \ \ddot{x} [/tex]

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

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.

[tex] - \ m \ g \ \frac{x}{l} \ = \ m \ \ddot{x} [/tex]

Leading to:

[tex] \ddot{x} \ + \ \frac{g}{l} \ x \ = \ 0 [/tex]

From this the angular frequency is shown to be [tex] \omega \ = \ \sqrt{ \frac{g}{l}}

[/tex]

The time period is calculated to be [tex] T \ = \ \frac{2 \ \pi}{ \omega} \ = \ 2 \ \pi \ \sqrt{ \frac{l}{g} } [/tex]

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.

--------------------------------------------------

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 [tex] T \ = \ 2 \ \pi \ \sqrt{ \frac{l}{g} } [/tex] period of the entire system?

If not I could use [tex] K.E. \ + \ U_{grav} \ + \ U_{spring} [/tex] 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.

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