A Pendulum Has A Length (L) And Time Period (T) For Completing One Oscillation. Prove That: F=1/2(L)1/2?

[inkay0]

A pendulum has a length (L) and time period (T) for completing one oscillation. prove that: f=1/2(L)1/2?

[fiction-factory]

Your Answer...You may want to check that you have typed this correctly. Is the second 1/2 a power- ie a square root? Can you clarify what f represents please

[fiction-factory]

Is f the frequency?

[Peter Pedant]

well it isnt is it?

the usual formula is T = ( l/g)1/2
( sqr root of l over g ) and g is the accel due to gravity

f is one over T and so F would be one over ( l/g) 1/2

and you can do it either by applying newtons second law to a plumb bob on a string
or
dimensional analysis

f has dimension T minus 1

L has dimension L to the power one
g has dimension L1, T minus 2

and if you have no idea what this means then re read the chapter this question comes from

it is straightforward - we were doing it in 1966

[bhg481]

There used to be a 2pi thrown into it when I was a lad in 1960. I used to watch the lamps swinging in morning assembly and work out the length of the cable (assuming pi squared approx = 10).

[fiction-factory]

Yes, as well as PP's g there should indeed be pi in there too.
The formula for T is easily found in any book/website on pendulums (or is it pendula). Then f is simply 1/T