anonymous
  • anonymous
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
Physics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
write down the equation of motion for pendulum
anonymous
  • anonymous
|dw:1349777134567:dw|
anonymous
  • anonymous
why is the mg force not straight down?

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anonymous
  • anonymous
sorry...mg is straight down only
anonymous
  • anonymous
does it help u in finding the time period?
anonymous
  • anonymous
I want to know how the formula is derived...
anonymous
  • anonymous
I know write down the equation of motion along the tangent... m( a_t) = mg sin(theta) a_t---tangential acc
anonymous
  • anonymous
it's fine?
anonymous
  • anonymous
net force along tangent is mg sin (theta)
anonymous
  • anonymous
yes
anonymous
  • anonymous
now v_tangential= r ( angular vel)
anonymous
  • anonymous
differentiate it a_tangential= r(angular acc)
anonymous
  • anonymous
ok
anonymous
  • anonymous
r is the length of pendulum in this case
anonymous
  • anonymous
now substitute a_tangential into m( a_t) = mg sin(theta)
anonymous
  • anonymous
m(r*angular_acc) = m(sin(theta))
anonymous
  • anonymous
m ( l(angular acc))= mg sin(theta)....l=r= length of pendulum
anonymous
  • anonymous
now l(angular acc))= g sin(theta)
anonymous
  • anonymous
where did the l come from?
anonymous
  • anonymous
l=r=length of pendulum...pendulum is like bob is moving in a circle pivoted at centre
anonymous
  • anonymous
oh right... l = r. I forgot about that
anonymous
  • anonymous
now acceleration = g/l sin(theta)
anonymous
  • anonymous
yep
anonymous
  • anonymous
equation of SHM is a= - w^2 x then time period T= 2 pi/w
anonymous
  • anonymous
I havent learnt SHM yet. whats w?
anonymous
  • anonymous
w(omega)---angular frequency acceleration = g/l sin(theta) for small theta acceleration = g/l (theta) sin(theta)= theta
anonymous
  • anonymous
u know wave... omega= 2 pi (nu)
anonymous
  • anonymous
yep
anonymous
  • anonymous
so we say SHM is a periodic motion in which acceleration is directed towards a fixed point which is called equilibrium point.
anonymous
  • anonymous
ok. I know the basics of it
anonymous
  • anonymous
ok.. so acceleration = g/l (theta)
anonymous
  • anonymous
for small theta
anonymous
  • anonymous
yep
anonymous
  • anonymous
now compare this with a= - w^2 x where x is the displacement from fixed point
anonymous
  • anonymous
so for small angle x = theta
anonymous
  • anonymous
so acceleration and displacement are in opposite direction
anonymous
  • anonymous
yes seems logical
anonymous
  • anonymous
here theta is x.... for small angle sin(theta)= theta
anonymous
  • anonymous
so in the case of pendulum angular displacement and angular acceleration are in opposite direction
anonymous
  • anonymous
write six in the form a power series sin x= x - x^3/3! + x^5/5!-------
anonymous
  • anonymous
so for small x sin x =x
anonymous
  • anonymous
neglect higher order terms
anonymous
  • anonymous
aha
anonymous
  • anonymous
so we have acceleration =- g/l (theta)
anonymous
  • anonymous
yes
anonymous
  • anonymous
so u have w^2= g/l
anonymous
  • anonymous
so T= 2 pi/w
anonymous
  • anonymous
I forgot what w is equal to
anonymous
  • anonymous
w= angular frequency= 2 pi( frequency)
anonymous
  • anonymous
so T = 2pi/2pi?
anonymous
  • anonymous
T= (2 pi)/ (2 pi( frequency))
anonymous
  • anonymous
:)
anonymous
  • anonymous
so T= 1/ (frequency)
anonymous
  • anonymous
so where does the √ come in?
anonymous
  • anonymous
w^2= g/l so what will be w?
anonymous
  • anonymous
√(g/l)
anonymous
  • anonymous
so T=?
anonymous
  • anonymous
2π T=----- √(g/l)
anonymous
  • anonymous
so u got the time period.
anonymous
  • anonymous
=2π√(l/g)
anonymous
  • anonymous
:D
anonymous
  • anonymous
I see one problem though... this will only be accurate with small angles right?

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