How to Express Period as a Function of Length?

[DB]

im having a little trouble with this part of my lab report, i think ill leave the details out, its just some math trouble.

here is my equation for frequency v.s function of length so that f\propto L^{-\frac{1}{2}}:

f=5.6*L^{-\frac{1}{2}}

now the question is re-write this equation expressing the Period (T) as a function of length L. Period being seconds per osc. (freq being osc. per second.)

wat i now so far is f=\frac{1}{T}=5.6*L^{-\frac{1}{2}} beyond that, i not really sure wat the question is asking me to do...



i have 1 more quick question. am i right to say that: yes, a pendulum's amplitude has an affect on its frequency because it has more distance to accelerate due to gravity, therefore giving it a greater velocity and higher frequency?

thanks

 

[HallsofIvy]

im having a little trouble with this part of my lab report, i think ill leave the details out, its just some math trouble.

here is my equation for frequency v.s function of length so that f\propto L^{-\frac{1}{2}}:

f=5.6*L^{-\frac{1}{2}}

now the question is re-write this equation expressing the Period (T) as a function of length L. Period being seconds per osc. (freq being osc. per second.)

wat i now so far is f=\frac{1}{T}=5.6*L^{-\frac{1}{2}} beyond that, i not really sure wat the question is asking me to do...

Solve the equation! Solve \frac{1}{T}= 5.6*L^{-\frac{1}{2}} for T.

i have 1 more quick question. am i right to say that: yes, a pendulum's amplitude has an affect on its frequency because it has more distance to accelerate due to gravity, therefore giving it a greater velocity and higher frequency?

thanks

This is a lab report isn't it? What did your experiment show about this?
As far as your argument is concerned, if the amplitude is greater, then the pendulum comes down more, is accelerated more due to gravity, and so has a greater velocity (at the bottom) but, then, it has farther to go doesn't it? Which effect is greater? Again, the best answer to that, in a lab report anyway, is your experimental result!

I love math, but in physics it is the experimental result that wins every argument!

 

[DB]

This is a lab report isn't it? What did your experiment show about this?
As far as your argument is concerned, if the amplitude is greater, then the pendulum comes down more, is accelerated more due to gravity, and so has a greater velocity (at the bottom) but, then, it has farther to go doesn't it? Which effect is greater? Again, the best answer to that, in a lab report anyway, is your experimental result!

I love math, but in physics it is the experimental result that wins every argument!

lol ur right about it being a lab report but the title of the thread kinda gives it away! anyway, on our lab sheet it said if you have time try increasing the amplitude by 20 degrees and see the results. unfortunatly my group didnt have enough time. but with ur statement in mind i believe that there will be no major difference in frequency with a change in amplitude, due to the fact that with a greater velocity the pendulum will have more acceleration distance yet will be going against gravity for more distance aswell, therefore canceling each other out when compared to a lower amplitude. is this making sense? thnx 4 da help btw

maybe the mass of the pendulum has the effect on the frequency?

 

[DB]

okay for new equation,
\frac{1}{T}=\frac{5.6*L^-{\frac{1}{2}}}{1}
\frac{T*(5.6*L^-{\frac{1}{2}})}{5.6*L^-{\frac{1}{2}}=\frac{1}{5.6*L^-\frac{1}{2}}
so
T=0.18*L^\frac{1}{2}
that right?

 

[DB]

u might not be able to see my second step because i can't lol
this might work
\frac{T*(5.6*L^-\frac{1}{2})}{5.6*L^-\frac{1}{2}}=\frac{1}{5.6*L^-\frac{1}{2}}