Explaining Pendulum: Period & Length Proportionality

[warriorofrovac]

Why is the period of the pendulum proportional to the square root of the length?

 

[arildno]

Well, what physical parameters would you think might affect the period?
How can you combine these parameters in order to get the physical dimension of "time" (i.e, the dimension of the period)

 

[Tide]

The vertical displacement of a pendulum during its swing is approximately L \theta^2 / 2 (using the small angle approximation for the cosine). Multiply by m g to get for the available potential energy for its motion.

Also, the speed of the pendulum as it passes its lowest point will be about

v \sim \frac {L \theta}{T}

Now set

g L \theta^2 / 2 \sim v^2/2 \sim \frac {1}{2} \left(\frac {L \theta}{T}\right)^2

to find T \sim \sqrt {L/g}.

 

[warriorofrovac]

The vertical displacement of a pendulum during its swing is approximately L \theta^2 / 2 (using the small angle approximation for the cosine). Multiply by m g to get for the available potential energy for its motion.

Also, the speed of the pendulum as it passes its lowest point will be about

v \sim \frac {L \theta}{T}

Now set

g L \theta^2 / 2 \sim v^2/2 \sim \frac {1}{2} \left(\frac {L \theta}{T}\right)^2

to find T \sim \sqrt {L/g}.

erm..can i say..i have no idea what you just said..physics isn't my strong point

 

[Tide]

erm..can i say..i have no idea what you just said..physics isn't my strong point

Well, you did ask a physics question in a physics forum so ... ! :)

 

[warriorofrovac]

Okay, forget the why question

What if the simple forumla showing that the period of the pendulum proportional to the square root of the length?

 

[Tide]

Okay,

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

 

[warriorofrovac]

Ah-ha. I thought so. Well i knew there was a 2pi in there somewhere..
I don't need to say why its like that...thankfully.
Hopefully that should be all the help i need...