Damn physics...

[Venix]

Someone already posted the equation and gave you the exact length of pendulum you need, IIRC. Look up a bit.

 

[DrPizza]

Originally posted by: esun
The equation clearly shows there is no dependence on mass or deflection (although smaller deflection is better because of ignored factors). Just get a piece of string and hang it from a peg so it drops exactly 0.2482 meters (or as close as you can get to that) with a mass at the bottom (preferably something moderately heavy but small), then swing away.

That's because the equation is only an approximation.... the larger the angle, the larger the error, due to (I think) some of the potential energy of the swing being converted into rotational energy for the mass.... at a small angle the mass doesn't rotate much.

 

[Brutuskend]

The answer I came up with. 😉

 

[falias]

Originally posted by: Venix
Someone already posted the equation and gave you the exact length of pendulum you need, IIRC. Look up a bit.

Sorry, should have clarified, I meant length of time for a swing (you said all pendulum periods are the same, regardless of where you release it), not length of the arm.

 

[mugs]

Originally posted by: falias

Originally posted by: esun
The equation clearly shows there is no dependence on mass or deflection (although smaller deflection is better because of ignored factors). Just get a piece of string and hang it from a peg so it drops exactly 0.2482 meters (or as close as you can get to that) with a mass at the bottom (preferably something moderately heavy but small), then swing away.

So you're saying speed is not a factor, or isn't that the acceleration of gravity (g)? I would have though where you release it for it to swing would matter how fast it completes a full swing, can't just swing away.....or am I wrong?

Yes you're wrong. It'll take the same amount of time to swing back and forth no matter where you release it. It will move faster the farther you pull it back, but it has a longer distance to travel. 🙂

 

[falias]

Originally posted by: Brutuskend
The answer I came up with. 😉

Thanks, I'll get right on it!
😛

 

[dighn]

Originally posted by: DrPizza

Originally posted by: esun
The equation clearly shows there is no dependence on mass or deflection (although smaller deflection is better because of ignored factors). Just get a piece of string and hang it from a peg so it drops exactly 0.2482 meters (or as close as you can get to that) with a mass at the bottom (preferably something moderately heavy but small), then swing away.

That's because the equation is only an approximation.... the larger the angle, the larger the error, due to (I think) some of the potential energy of the swing being converted into rotational energy for the mass.... at a small angle the mass doesn't rotate much.

not sure about that rotation thing but that equation uses the approximation of sin(a) ~= a at small angles, i found this through google: http://www.vislab.usyd.edu.au/education/chaos/idjChaosLAB/node17.html

 

[falias]

Originally posted by: mugsywwiii

Originally posted by: falias

Originally posted by: esun
The equation clearly shows there is no dependence on mass or deflection (although smaller deflection is better because of ignored factors). Just get a piece of string and hang it from a peg so it drops exactly 0.2482 meters (or as close as you can get to that) with a mass at the bottom (preferably something moderately heavy but small), then swing away.

So you're saying speed is not a factor, or isn't that the acceleration of gravity (g)? I would have though where you release it for it to swing would matter how fast it completes a full swing, can't just swing away.....or am I wrong?

Yes you're wrong. It'll take the same amount of time to swing back and forth no matter where you release it. It will move faster the farther you pull it back, but it has a longer distance to travel. 🙂

Makes sense, so that must mean there is a standard time for a pendulum to make a full swing, I wanted to know what that was...

 

[Joker81]

I found this as the first search. You just need to be able to make a spring clock. Suprisingly Pendulum clocks didn't arrive until the 1650's


Clocks From an earlier age

You could just find a stream or somthing and make a wheel that spins with the water and just count the number of times it rotates. It would be somewhat acurate i don't know if 1 second per 30 minutes.


Edit. Find some radioactive materials that decays pretty rapidly. Weigh it before and weigh it after. And propose that the radioactive element did exist in that time.

 

[RossGr]

Originally posted by: falias

Originally posted by: mugsywwiii

Originally posted by: falias

Originally posted by: esun
The equation clearly shows there is no dependence on mass or deflection (although smaller deflection is better because of ignored factors). Just get a piece of string and hang it from a peg so it drops exactly 0.2482 meters (or as close as you can get to that) with a mass at the bottom (preferably something moderately heavy but small), then swing away.

So you're saying speed is not a factor, or isn't that the acceleration of gravity (g)? I would have though where you release it for it to swing would matter how fast it completes a full swing, can't just swing away.....or am I wrong?

Yes you're wrong. It'll take the same amount of time to swing back and forth no matter where you release it. It will move faster the farther you pull it back, but it has a longer distance to travel. 🙂

Makes sense, so that must mean there is a standard time for a pendulum to make a full swing, I wanted to know what that was...

There is no "standard" time for a pendulum swing, the time for a "swing" depends on the length, so you can control the time for a "swing" by changing the length. Many old clocks adjusted their rate by moving the end weights up or down to effect the rate of osscilation. Thus if you can design an adjustable weight for the end of you pendulumn you will be able to fine tune the period.