How did scientists compose the equation for period?

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The equation for the period of a pendulum, T=2π√(l/g), is derived from the principles of simple harmonic motion (SHM). The units of √(l/g) simplify to seconds, while 2π represents radians per revolution. The discussion clarifies that the negative sign in the force equation, F=-mgsinθ, arises from the gravitational force acting on the pendulum at an angle. The relationship between angular frequency (ω) and acceleration is established, leading to the identification of ω² as g/l. Understanding these concepts is essential for grasping the derivation of the pendulum's 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 ?
 
  • #10
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.
 
  • #11
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:

T=2 \pi \sqrt{\frac{l}{g}}


this is the simplest way to show it IMO
 
  • #12
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 :)
 
  • #13
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?
 
  • #14
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
 
  • #15
What's the x stand for?
 
  • #16
kashiark said:
What's the x stand for?

at the angle θ, the string marks out an arc length x
 
  • #17
I get it! Thanks!
 

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