Period of a pendulum

  • Thread starter zeralda21
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  • #1
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Homework Statement



For a mathematical pendulum, you notice that if you extend the string with 60 cm, then the time of one period will double for small oscillations. What was the period of time before the line was extended?





Homework Equations



T=2pi*sqrt(l/g) where l is the lenght of the pendulum.


The Attempt at a Solution



If I assume that the first pendulum has a lenght l, then the period will of course be: T=2pi*sqrt(l/g). If I now extend it by 0.6m, the period will double: 2T=2pi*sqrt((l+0.6)/g) which is equal to T=pi*sqrt((l+0.6)/g).

Now I set my first equation equal to the last one:

T=2pi*sqrt(l/g)=T=pi*sqrt((l+0.6)/g) , squaring LHS and RHS----> 4l/g=(l+0.6)/g ---->

4l=l+0.6---> l=5/6. If I insert this value of l into my first equation I get that T=1.83 seconds which is wrong. Correct answer is 0.9 seconds.

Or is there another way of solving?
 
Last edited:

Answers and Replies

  • #2
Doc Al
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Now I set my first equation equal to the last one:

T=2pi*sqrt(l/g)=T=pi*sqrt((l+0.6)/g) , squaring LHS and RHS----> 4l/g=(l+0.6)/g ---->
Looks OK.
4l=l+0.6---> l=5/6.
Redo that last step!
 
  • #3
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Looks OK.

Redo that last step!
What a mistake..I assumed 0.6=5/3 instead of 3/5. The lenght 1/5 is correct and matches with period. Thank you very much sir.
 

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