How did scientists compose the equation for period?

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T=2pi√(l/g)

Is it in radian*seconds?
 

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Hi Kashiark-

What are the units of sqrt(L/g)? of sqrt(g/L)?

Bob S
 
√(l/g) would be √m/(√m/s) or √m*(s/√m) = s, and I was just guessing that the 2pi was radians; isn't it?
 
Err...no. Sqrt(L/g) is in units of seconds per radian, snd sqrt(g/L) is in units of radians per second. 2 pi has units of radians per revolution.
 
Why is √(l/g) in units of seconds per radian? Length is in meters, and g is in m/s². Where did the radians come from?
 
Pi have no physical units. The radian is defined as a ratio (length of the arc/radius).
It's just a matter of convention to say that angular speed is in radians/s. The actual physical unit for angular speed is just 1/s.
Look at v=omega*r
v is not in m*rad/s, is it?
 
Ok, that makes sense, but why is there a 2pi in the equation? What was the argument made when someone said, "Hey, I think period equals 2 pi times the square root of length over the acceleration due to gravity!"
 
kashiark said:
Ok, that makes sense, but why is there a 2pi in the equation? What was the argument made when someone said, "Hey, I think period equals 2 pi times the square root of length over the acceleration due to gravity!"

you should read about the http://en.wikipedia.org/wiki/Pendulum_(mathematics)" then.
 
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Wow... I'm kind of overwhelmed. I can follow it from s=lΘ to d²Θ/dt²+(g/l)sinΘ=0 ... From there I'm too ignorant about differential equations to follow it, but why does F=-mgsinΘ ?
 
kashiark said:
Wow... I'm kind of overwhelmed. I can follow it from s=lΘ to d²Θ/dt²+(g/l)sinΘ=0 ... From there I'm too ignorant about differential equations to follow it, but why does F=-mgsinΘ ?

There's a vector diagram to the right of that equation which explains it pretty clearly. Gravity exerts -mg on the pendulum; the component parallel to the pendulum's motion is -mgsinΘ.
 
kashiark said:
Wow... I'm kind of overwhelmed. I can follow it from s=lΘ to d²Θ/dt²+(g/l)sinΘ=0 ... From there I'm too ignorant about differential equations to follow it, but why does F=-mgsinΘ ?

The force mg acts at an angle θ, so the components are one tangentially mgsinθ and the other mgcosθ. mgsinθ points in the opposite direction of motion so it is -mgsinθ

that is how F= -mgsinθ and we know F=ma

so ma = -mgsinθ => a= -gsinθ for small θ, sinθ≈θ so a=-gθ. θ=x/l (arc length/radius)

so a = -(g/l)x

which is of the form a= -ω2x, so it exhibits SHM, for ω2=g/l. And for SHM T=2π/ ω which works out as:

[tex]T=2 \pi \sqrt{\frac{l}{g}}[/tex]


this is the simplest way to show it IMO
 
What does tangentially mean? I'm familiar with the function itself, but I'm not sure what you mean by it. After there, I can follow you until a=-ω²x; where did the ω come from? I have another problem after there, but we'll get to that later I suppose if you guys still want to help me :)
 
Forget the first part, I just got that from thinking about it while brushing my teeth, but I still don't understand why it should be a negative mgsinθ
Edit: Ok, I figured that out as well, but I'm still stuck with the ω. It wasn't in any equation one second, and then it was. What does it represent anyway?
 
kashiark said:
Forget the first part, I just got that from thinking about it while brushing my teeth, but I still don't understand why it should be a negative mgsinθ
Edit: Ok, I figured that out as well, but I'm still stuck with the ω. It wasn't in any equation one second, and then it was. What does it represent anyway?

angular frequency
 
What's the x stand for?