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

1. Michele_Laino

hint: we have: $\sqrt {\frac{l}{g}} = \sqrt {\frac{{{\text{meters}}}}{{{\text{meters/second}}{{\text{s}}^{\text{2}}}}}}$

2. TrojanPoem

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

it is correct! namely option a)

4. TrojanPoem

why ?

5. TrojanPoem

Ops

6. Michele_Laino

a good method is to start from the differential equation of the motion of the pendulum

7. TrojanPoem

m/m/s^2 = sqrt(s^2)

8. TrojanPoem

But why b is wrong ?

9. Michele_Laino

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

we cannot check if 2 pi is correct therefore b is correct

11. Michele_Laino

in other words, using the dimensional analysis I am able to write this: $T \propto \sqrt {\frac{l}{g}}$

12. TrojanPoem

Yeah, so correct according to dimension ?

13. Michele_Laino

no, since you have this formula: $T = 2\pi \sqrt {\frac{l}{g}}$

14. TrojanPoem

a or b ?

15. Michele_Laino

I think it is option a)

16. TrojanPoem

Doesn't a mean "We are fully sure that the equation dimensions are correct, constants are correct" ?

17. Michele_Laino

this formula: $T \propto \sqrt {\frac{l}{g}}$ is correct according to dimensions

18. TrojanPoem

so if the constant is not written it's correct according to dimensions if it's written , just correct ?

19. Michele_Laino

yes!

20. TrojanPoem

THANKS.