Scaling factor of a simple pendulum between length and time period.

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SUMMARY

The time period of a simple pendulum is directly proportional to the square root of its length, as described by the equation T = 2π√(l/g). In this discussion, it is established that when the length of the pendulum is increased by 3.0m, the time period doubles, leading to the conclusion that the original length of the pendulum is 1m. The mathematical manipulation involved correctly incorporates the factor of 2 into the square root, resulting in a new length that is four times the original length.

PREREQUISITES
  • Understanding of simple harmonic motion
  • Familiarity with the formula for the time period of a pendulum (T = 2π√(l/g))
  • Basic algebraic manipulation skills
  • Knowledge of proportional relationships in physics
NEXT STEPS
  • Explore the effects of varying mass on the time period of a pendulum
  • Learn about the impact of gravitational acceleration on pendulum motion
  • Investigate the relationship between pendulum length and frequency
  • Study the mathematical derivation of the time period for different pendulum shapes
USEFUL FOR

Students studying physics, educators teaching mechanics, and anyone interested in the principles of pendulum motion and harmonic oscillators.

mrcotton
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Homework Statement



The time period of a simple pendulum is doubled when the length of the pendulum is increased by 3.0m. What is the original length of the pendulum?

Homework Equations



T= 2∏√(l/g)
also l is original length and l+3 is the new length

The Attempt at a Solution



So
2T=2(2∏√(l/g)
2T=2∏√(4l/g)
The new length is 4 times the original length and it also has to be the original length plus 3m.
∴ 4l=l+3
so l=1m

I am happy with the mathematical steps, and it gives me the correct answer.
It is the reason why I can do this I could do with some guidance on.
Is it because when I double T the only thing that could change is the length?
Therefore I can put the multiplying factor of 2 inside the sqrt as 4×l.
 
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hi mrcotton! :wink:
mrcotton said:
It is the reason why I can do this I could do with some guidance on.
Is it because when I double T the only thing that could change is the length?
Therefore I can put the multiplying factor of 2 inside the sqrt as 4×l.

yes, that's exactly right …

T = 2π√(l/g) means that T is proportional to √l, so twice T means 4 times l (if everything else is the same) :smile:

(and your proof is completely correct)
 
What you did is fine. But you were lucky to have been able to sneak the "2" inside the square root sign so easily to make 4.

I would have said

T = 2π√(l/g)
2T = 2π√[(l+3)/g]
so 2 = √[(l+3)/l]
& solve for l.
 
isn't that the same?
 
Thanks ever so much for the help with getting my old grey matter working.
So in general I can solve it like this.
Q85_zps9e5c662b.jpg

Did the sqrt of 4 only work for those values.
I tried doing the calculation both ways for a pendulum that's time period has tripled when the original length is extended by 6m and got the same results.
 
Yeah, maybe I was a bit severe here ... did you try for new l = 1.73l? This is a very trivial point, should probably not even have raised it.
 
Hi Rude man, please I like the trivial points. Its the pedantic points in physics that make it interesting to me. Each physics question seems to open up more questions.
 
mrcotton said:
Hi Rude man, please I like the trivial points. Its the pedantic points in physics that make it interesting to me. Each physics question seems to open up more questions.

OK I got plenty of those ... :smile:
 

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