Predict the equation for the period of a pendulum

  • #1
I did a lab on pendulums and I need to answer the following:
Examine the experimental evidence in regards to each of the properties of the pendulum, mass, horizontal displacement and length. Predict the equation for the period of a pendulum and calculate it based on your observations.

The data I have to work with is:
-I have different horizontal displacements and the periods for each displacement
-I have different masses and the periods for each mass
-I have different pendulum lengths and the periods for each length

In this lab, I observed that neither the horizontal displacement or the mass had an effect on the period of a pendulum.


I found this page ( http://hyperphysics.phy-astr.gsu.edu/Hbase/pend.html#c4 ) that works out the steps for deriving the period of a pendulum, but I, nor any classmates of mine, have been introduced to calculus yet so I am assuming that is not the way I am supossed to derive this equation.

I graphed out the period squared vs the pendulum length and they are proportional.

If anyone could help with with some sort of direction to go in creating the equation of a pendulum I would really appreciate it!
Thanks (:
 

Answers and Replies

  • #2
19
0


So I am guessing that you are asking for the equation of motion for the pendulum.

Start out with the first law for torque

[tex] T = I\alpha [/tex]

For the pendulum the moment of intertia is

[tex] I = L^2m [/tex]

So what are the torques on the pendulum? If you draw out the free body diagram (or look on the site) and use a little trignometry you find that there is one due to gravity

[tex] T = -mgL\sin\theta [/tex]

Combining the two equations gives you

[tex] L^2m\alpha = -mgL\sin\theta [/tex]

Now here is the one part where a very little bit of calculus comes in. Acceleration is the second derivative of displacement with respect to time. This can be written

[tex] \alpha = \frac{d^2\theta}{dt} [/tex]

Another relation you need to use is that for small angles the following is true

[tex] \sin\theta = \theta [/tex]

Combing the previous three equations and rearranging gives

[tex] \frac{d^2\theta}{dt} + \frac{g}{L}\theta = 0 [/tex]

Which is the equation of motion for the pendulum.
 
  • #3
Delphi51
Homework Helper
3,407
10


I graphed out the period squared vs the pendulum length and they are proportional.
Then you have the formula: T² = kL
(the quantity on the vertical axis = slope x quantity on horizontal axis)
It would be nice to express the slope constant in terms of g, but that would be working backward from the answer. Unless you have the means to do the experiment under different g conditions!
 
  • #4


I think I may have been unclear in my explanation, I THINK I am trying to get to the equation:
b3129742fed41e2f6d2e6962c78a3cdc.png

Is there any way I can get there from T^2=kL?
Or am I completely in the wrong direction?

Thanks for the replies (:
 
  • #5
Delphi51
Homework Helper
3,407
10


Take the square root of both sides.
You might compare your slope value with the 4π²/g from the formula and find the % difference.

Mbaboy, that is the nicest derivation of the formula I've seen! Thank you very much.
 
  • #6


Take the square root of both sides.
You might compare your slope value with the 4π²/g from the formula and find the % difference.

Mbaboy, that is the nicest derivation of the formula I've seen! Thank you very much.
I think that will work out quite nicely!
thank you very much for your help!
 

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