# Hi please explaint it step by steps. ThanksIt can be shown that the exact ode for the pendulum problem is d2 Ɵ/dt2+ g/Lsin Ɵ = 0. By leeting d2 Ɵ/dt2= wdw/d Ɵ, where w is the angular...

Hi please explaint it step by steps.

Thanks

*It can be shown that the exact ode for the pendulum problem is d2** Ɵ/dt2+ g/Lsin Ɵ = 0.*

*By leeting d2** Ɵ/dt2= wdw/d Ɵ, where w is the angular velocity, solvethe ode to obtain w in terms of Ɵ .*

*L = 0.5m , `Theta` = π/3, g = 9.81ms-2 and compare your result with question 5 using the maples evalf command to evaluate the integral.*

*print*Print*list*Cite

### 1 Answer

You need to replace `(d^2 Theta)/(dt^2)` by `(wdw)/(dTheta) ` in given ode such that:

`(wdw)/(dTheta) + (g/L)*sin Theta = 0`

You need to move `(g/L)*sin Theta` to the right side such that:

`(wdw)/(dTheta)= -(g/L)*sin Theta`

`` You need to separate the variables such that:

`wdw = -(g/L)*sin Theta dTheta`

You need to integrate both sides such that:

`int wdw = int -(g/L)*sin Theta dTheta`

You need to factor out the constant term `-(g/L)` such that:

`w^2/2 = -(g/L)*(-cos Theta) + c`

`` `w^2 = 2(g/L)*(cos Theta) + c`

`w = +- sqrt(2(g/L)*(cos Theta))+c`

You need to substitute g by `9.8m/(s^2), ` L by`0.5 m` and `Theta` by `pi/3` , such that:

`w = +-sqrt(2*(9.8/0.5)*cos (pi/3))`

`w = +-sqrt (2*19.6*(1/2)) =gt w~~+-4.427 (rad)/s`

**Hence, evaluating the angular velocity for the given values g,L,`Theta` , yields `w~~+-4.427 (rad)/s` .**