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Name: Solve d2θdt2+gsin⁡θ=0
Uploaded: 2022-02-23T17:22:57+00:00
Description: Solve d2θdt2+gsin⁡θ=0

Mary Keefe

Answered question

2022-01-19

Solve

\frac{d^{2} θ}{{d t}^{2}} + g \sin θ = 0

Answer & Explanation

Piosellisf

Beginner2022-01-19Added 40 answers

Use substitution:

θ^{'} = v

, therefore we have that:

θ^{″} = \frac{d v}{d t} \cdot \frac{d t}{d θ} \cdot \frac{d θ}{d t} \Rightarrow θ^{″} = \frac{d v}{d θ} \cdot v \Rightarrow θ^{″} = v^{'} \cdot v

where v is function in terms of variable

θ

. So differential equation becomes:

v^{'} \cdot v + g \cdot \sin θ = 0

which is separable differential equation.

Mary Goodson

Beginner2022-01-20Added 37 answers

Start with

\frac{1}{2} \frac{d {\dot{θ}}^{2}}{d θ} = θ \frac{d \dot{θ}}{d θ} = \frac{d θ}{d t} \frac{d \dot{θ}}{d θ} = \frac{d \dot{θ}}{d t} = \ddot{θ}

Then your equation becomes

\frac{1}{2} \frac{d {\dot{θ}}^{2}}{d θ} = - \frac{g}{l} \sin (θ)

d {\dot{θ}}^{2} = - \frac{2 g}{l} d \sin (θ) \Rightarrow {\dot{θ}}^{2} = \frac{2 g}{l} \cos (θ) + c_{1}

It's a bit easier if we assume initial conditions, say

\dot{θ} (t_{0}) = {\dot{θ}}_{0} and θ (t_{0}) = θ_{0}

, so that

{\dot{θ}}^{2} = \frac{2 g}{l} [\cos (θ) - \cos (θ_{0}) + \frac{l {\dot{θ}}_{0}^{2}}{2 g}]

Then

\frac{d θ}{d t} = \sqrt{\frac{2 g}{l}} \sqrt{\cos (θ) - \cos (θ_{0}) + \frac{l {\dot{θ}}_{0}^{2}}{2 g}}

so that

d t = \sqrt{\frac{l}{2 g}} \frac{d θ}{\sqrt{\cos (θ) - \cos (θ_{0}) + \frac{l {\dot{θ}}_{0}^{2}}{2 g}}}

t_{f} - t_{0} = \sqrt{\frac{l}{2 g}} \int_{θ_{0}}^{θ_{f}} \frac{d θ}{\sqrt{\cos (θ) - \cos (θ_{0}) + \frac{l θ_{0}^{2}}{2 g}}}

This equation is of the form

t = f (θ)

. Your solution is given by

θ = f^{- 1} (t)

. That's about as much as you need to know, since it's more efficient to just solve the original equation numerically.

RizerMix

Expert2022-01-27Added 656 answers

replacing

\sin θ by θ

(physically assuming small angle deflection) gives you a homogeneous second order linear differential equation with constant coefficients, whose general solution can be found in most introductory diff eq texts (or a google search). this new equation represents a simple harmonic oscillator (acceleration proportional to displacement, like a spring force).

θ + g θ = 0

has solution

A \cos (\sqrt{g} t) + B \sin (\sqrt{g} t)

. so, for example, if the initial displacement is

θ_{0}

and initial angular velocity is 0 then the solution is

θ_{0} \cos (\sqrt{g} t)

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