pendulum

[Tetraspace]

i need help on pendulums on the variable that affects the swing time of it. thnaks

[YT2095]

the distance of the bob from the pivot/hinge

 

more subtle things like air resistance and pivot friction apply too, but only for very accurate measurments.

 

that`s the only thing that spring to mind at the moment

 

edit: as an afterthought, it needn`t necesarily be air resistance either, Air was just an example.

[swansont]

The period depends on sqrt (l/g)

 

So, as YT said, the distance to the bob, which can depend on environmental factors such as temperature, as well as the accceleration due to gravity. All else being equal, a pendulum swings more slowly on the top of a mountain, or even when the moon is overhead.

[Phospholipid]

The period depends on sqrt (l/g)

 

So' date=' as YT said, the distance to the bob, which can depend on environmental factors such as temperature, as well as the accceleration due to gravity. All else being equal, a pendulum swings more slowly on the top of a mountain, or even when the moon is overhead.[/quote']

 

Actually, the period depends on 2 x pi x sqrt (l/g)

 

Also, Any kind of torque will affect a pendulum, as well as movement into another plane (which means the mass at the bottom of the pendulum did not reach its maximum height as has both a vertical and a horizontal compenent.. so the time it takes to swing would be a little shorter than expected.)

 

With that being said, I too, have a small question:

 

Why should and angle through which the pendulum swings be no greater than 10 degrees... and what happens as the angle becomes larger?

[swansont]

Actually' date=' the period depends on 2 x pi x sqrt (l/g)

 

Also, Any kind of torque will affect a pendulum, as well as movement into another plane (which means the mass at the bottom of the pendulum did not reach its maximum height as has both a vertical and a horizontal compenent.. so the time it takes to swing would be a little shorter than expected.)

 

With that being said, I too, have a small question:

 

Why should and angle through which the pendulum swings be no greater than 10 degrees... and what happens as the angle becomes larger?[/quote']

 

Since pi isn't a variable I ignored it. We are interested in fluctuations - things that can change the period.

 

The reason to use small angles is that the derivation uses the small-angle approximation of sin theta = theta (using radians), but this approximation gets worse as the angle gets larger. You'd have to go through the derivation to see where the error term shows up - I don't recall off the top of my head.

[The Rebel]

Its from the formula Torque = rotational inertia x angular acceleration.

 

Picture the pendulum at a slight angle, the force acting down is mg, so the force along the length is mg cos theta, and consequently the torque force is -mg sin theta. So, [math]T = l * -mg sin \theta[/math]

 

Angular acceleration is simply the second derivative of the angular displacement [math]\frac{d^2\theta}{dt^2}[/math]

 

Rotational Inertia = [math]ml^2[/math]

 

So . . . [math]lmg sin \theta=-ml^2\frac{d^2\theta}{dt^2}[/math]

 

which rearranged becomes [math]\frac{d^2\theta}{dt^2}=-\frac{l}{g}sin\theta[/math]

 

Simple Harmonic Motion is defined by [math]a=-w^2x[/math]

 

If we assumed the angle to be small, [math]sin\theta=\theta[/math], the pendulum would obey SHM, otherwise it doesn't.

 

An example would be if we increased the angle to [math]\pi[/math]. The pendulum would just drop and not obey any form of SHM. (unless the length is a spring)

[swansont]

If we assumed the angle to be small' date=' [math']sin\theta=\theta[/math], the pendulum would obey SHM, otherwise it doesn't.

 

Right. The McLauren expansion with the next term is ^{[math]sin\theta=\theta -\theta^3 /6 + ...[/math]}

 

so as long as ^{[math]\theta^3 /6 [/math]} is negligible, you're OK. Otherwise you have to include that and possibly higher-order terms, and then you can't solve the diff eq easily.