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TrojanPoem

  • one year ago

Verify if this equation is true based on physics dimensions : T = 2 pi sqrt(l/g) As T = time, l = length , g = gravity

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  1. Michele_Laino
    • one year ago
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    hint: we have: \[\sqrt {\frac{l}{g}} = \sqrt {\frac{{{\text{meters}}}}{{{\text{meters/second}}{{\text{s}}^{\text{2}}}}}} \]

  2. TrojanPoem
    • one year ago
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    I've solved it and got T^-1 as the dimension. The problem is , I can't find which answer is correct a) correct b) correct according to dimensions c) wrong d) wrong according to dimensions d or c ? and why ?

  3. Michele_Laino
    • one year ago
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    it is correct! namely option a)

  4. TrojanPoem
    • one year ago
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    why ?

  5. TrojanPoem
    • one year ago
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    Ops

  6. Michele_Laino
    • one year ago
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    a good method is to start from the differential equation of the motion of the pendulum

  7. TrojanPoem
    • one year ago
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    m/m/s^2 = sqrt(s^2)

  8. TrojanPoem
    • one year ago
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    But why b is wrong ?

  9. Michele_Laino
    • one year ago
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    yes! it is correct, nevertheless your reasoning is not able to justify the presence of the coefficient \(2 \pi\) option b) is wrong since we have the correct formula, and the dimensional analysis can not say anything on possible numerical coefficients which may accompany the various formulas

  10. TrojanPoem
    • one year ago
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    we cannot check if 2 pi is correct therefore b is correct

  11. Michele_Laino
    • one year ago
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    in other words, using the dimensional analysis I am able to write this: \[T \propto \sqrt {\frac{l}{g}} \]

  12. TrojanPoem
    • one year ago
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    Yeah, so correct according to dimension ?

  13. Michele_Laino
    • one year ago
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    no, since you have this formula: \[T = 2\pi \sqrt {\frac{l}{g}} \]

  14. TrojanPoem
    • one year ago
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    a or b ?

  15. Michele_Laino
    • one year ago
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    I think it is option a)

  16. TrojanPoem
    • one year ago
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    Doesn't a mean "We are fully sure that the equation dimensions are correct, constants are correct" ?

  17. Michele_Laino
    • one year ago
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    this formula: \[T \propto \sqrt {\frac{l}{g}} \] is correct according to dimensions

  18. TrojanPoem
    • one year ago
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    so if the constant is not written it's correct according to dimensions if it's written , just correct ?

  19. Michele_Laino
    • one year ago
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    yes!

  20. TrojanPoem
    • one year ago
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    THANKS.

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