## apple_pi 3 years ago Where does the equation for simple pendulum motion come from? $T=2\pi \sqrt{\frac{ l }{ g }}$ where T is period, l is string length, g is acc. due to gravity

1. akash123

write down the equation of motion for pendulum

2. akash123

|dw:1349777134567:dw|

3. apple_pi

why is the mg force not straight down?

4. akash123

sorry...mg is straight down only

5. akash123

does it help u in finding the time period?

6. apple_pi

I want to know how the formula is derived...

7. akash123

I know write down the equation of motion along the tangent... m( a_t) = mg sin(theta) a_t---tangential acc

8. akash123

it's fine?

9. akash123

net force along tangent is mg sin (theta)

10. apple_pi

yes

11. akash123

now v_tangential= r ( angular vel)

12. akash123

differentiate it a_tangential= r(angular acc)

13. apple_pi

ok

14. akash123

r is the length of pendulum in this case

15. akash123

now substitute a_tangential into m( a_t) = mg sin(theta)

16. apple_pi

m(r*angular_acc) = m(sin(theta))

17. akash123

m ( l(angular acc))= mg sin(theta)....l=r= length of pendulum

18. akash123

now l(angular acc))= g sin(theta)

19. apple_pi

where did the l come from?

20. akash123

l=r=length of pendulum...pendulum is like bob is moving in a circle pivoted at centre

21. apple_pi

oh right... l = r. I forgot about that

22. akash123

now acceleration = g/l sin(theta)

23. apple_pi

yep

24. akash123

equation of SHM is a= - w^2 x then time period T= 2 pi/w

25. apple_pi

I havent learnt SHM yet. whats w?

26. akash123

w(omega)---angular frequency acceleration = g/l sin(theta) for small theta acceleration = g/l (theta) sin(theta)= theta

27. akash123

u know wave... omega= 2 pi (nu)

28. apple_pi

yep

29. akash123

so we say SHM is a periodic motion in which acceleration is directed towards a fixed point which is called equilibrium point.

30. apple_pi

ok. I know the basics of it

31. akash123

ok.. so acceleration = g/l (theta)

32. akash123

for small theta

33. apple_pi

yep

34. akash123

now compare this with a= - w^2 x where x is the displacement from fixed point

35. apple_pi

so for small angle x = theta

36. akash123

so acceleration and displacement are in opposite direction

37. apple_pi

yes seems logical

38. akash123

here theta is x.... for small angle sin(theta)= theta

39. akash123

so in the case of pendulum angular displacement and angular acceleration are in opposite direction

40. akash123

write six in the form a power series sin x= x - x^3/3! + x^5/5!-------

41. akash123

so for small x sin x =x

42. akash123

neglect higher order terms

43. apple_pi

aha

44. akash123

so we have acceleration =- g/l (theta)

45. apple_pi

yes

46. akash123

so u have w^2= g/l

47. akash123

so T= 2 pi/w

48. apple_pi

I forgot what w is equal to

49. akash123

w= angular frequency= 2 pi( frequency)

50. apple_pi

so T = 2pi/2pi?

51. akash123

T= (2 pi)/ (2 pi( frequency))

52. akash123

:)

53. akash123

so T= 1/ (frequency)

54. apple_pi

so where does the √ come in?

55. akash123

w^2= g/l so what will be w?

56. apple_pi

√(g/l)

57. akash123

so T=?

58. apple_pi

2π T=----- √(g/l)

59. akash123

so u got the time period.

60. apple_pi

=2π√(l/g)

61. apple_pi

:D

62. apple_pi

I see one problem though... this will only be accurate with small angles right?