Question

Solve: `d^2x//dt^2 + g sin theta t =0` if `theta=g//l` , and g and l are constants?

Answer 1

`x(t) = - g sin(g/1) t^2/2 + At + B`

Explanation:

You are asking for the solution to:

`(d^2x)/(dt^2) = - g sin(g/1)`

`implies (d x)/(dt) = - g sin(g/1) t + A`

`implies x(t) = - g sin(g/1) t^2/2 + At + B`

Answer 2

`x = l^2/g \ sin theta t + At + B`

Explanation:

We have:

`(d^2x)/(dt^2) + g sintheta t = 0` where `theta=g/l`, a constant

Which we can write as:

`(d^2x)/(dt^2) = - g sintheta t`

We can "separate the variables" , to get:

`(dx)/(dt) = int \ - g sintheta t \ dt`

And integrating give us:

`(dx)/(dt) = g/theta cos theta t + A`

And repeating we get:

`x = int \ g/theta cos theta t + A \ dt`

So that:

`x = g/theta^2 sin theta t + At + B`

`\ \ = g/(g/l)^2 sin theta t + At + B`

`\ \ = g l^2/g^2 \ sin theta t + At + B`

`\ \ = l^2/g \ sin theta t + At + B`