Solving a Second-Order Differential Equation

In summary, the conversation discusses the difficulty in solving a differential equation for the period of a physical pendulum. The equation is non-linear and cannot be solved in a simple form. Different techniques, such as making the small-angle approximation or using perturbation theory, can be used to find approximate solutions. It is also mentioned that there are tables of values for solving elliptic integrals, which are necessary for finding a theoretical formulation for large-angle values. The conversation ends with a question about whether there has been a series solution for the exact motion of an undamped pendulum.
  • #1
Saketh
261
2
I've been trying to derive the period of a physical pendulum, and I ended up with a differential equation that boils down to this:

[tex]
\frac{d^2 \theta}{dt^2} = C \sin{\theta}
[/tex]

where C is some constant.

With no experience in differential equations, I have no idea how to solve this. I can't find anything about how to solve this type of differential equation, so if someone could point me in the right direction, that would be helpful.

Thanks!
 
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  • #2
It can't be solved nicely, which is why we usually assume the angle is small and make the approximation [itex] \sin \theta \approx \theta[/itex]. If you want, there's a way to reduce it to finding the integral of something like [itex]1/\sqrt{a-\cos(\theta)}[/tex], but that's not a nice integral to do (ie, you need special functions).
 
  • #3
That is an extremely non-linear equation and cannot be solved in any simple form. As StatusX said, for smal values of the angle, you can make the approximation [itex]sin(\theta)= \theta[/itex] so the equation becomes [itex]d^2\theta/dt^2= C\theta[/itex].

Another technique is called "quadrature". Since t does not appear explictly in the equation, let [itex]\omega= d\theta/dt[/itex]. Then apply the chain rule: [itex]d^2\theta/dt^2= d\omega/dt= d\omega/d\theta d\theta/dt= \omega d\omega/dt= Csin \theta[/itex]. That's a separable first order differential equation which can be written as [itex]\omega d\omega= Csin(\theta)d\theta[/itex] and can be integrated directly:
[itex](1/2) \omega^2= -Ccos(\theta)+ D[/itex]. The rub is when you replace [itex]\omega[/itex] by [itex]d/t\eta/dt[/itex] and solve for [itex]d\theta/dt[/itex]: [itex] d\theta/dt= \sqrt{D -2C cos\theta}[/itex] so
[itex]\frac{d\theta}{\sqrt{D- 2C cos(\theta}}= dt[/itex] and the left side cannot be integrated in terms of any elementary function. (That's called an "elliptic" integral. I've seen whole book cases of tables of values.)
 
  • #4
Does this mean that it is impossible to generate a theoretical formulation for pendulums that includes large-angle values?

I suspected that I would have to make a small-angle assumption. If I did have [itex]d^2\theta/dt^2= C\theta[/itex], how would I solve that? (Sorry for my lack of experience - once again, I just need to be pointed in the right diretcion.)
 
  • #5
For small angles, yes. Of course, in order to get periodic solutions, C must be negative.
 
  • #6
what function equals itself times a constant when differentiating twice?
(HINT: what's the derivative of e^x ?)
 
  • #8
Saketh said:
Does this mean that it is impossible to generate a theoretical formulation for pendulums that includes large-angle values?

I suspected that I would have to make a small-angle assumption. If I did have [itex]d^2\theta/dt^2= C\theta[/itex], how would I solve that? (Sorry for my lack of experience - once again, I just need to be pointed in the right diretcion.)
You can get an approximate solution to this, to whichever degree of accuracy you desire, by perturbation theory.

Alternatively, you may use numerics
 
  • #9
daniel_i_l said:
what function equals itself times a constant when differentiating twice?
(HINT: what's the derivative of e^x ?)
I don't understand - how does that help me solve the differential equation?

Thanks, everyone! http://scienceworld.wolfram.com/physics/Pendulum.html" answers all of my questions.
 
Last edited by a moderator:
  • #10
daniel_i_l said:
what function equals itself times a constant when differentiating twice?
(HINT: what's the derivative of e^x ?)
I can think of 4 such (independent) functions. That's why I said " Of course, in order to get periodic solutions, C must be negative."
 
  • #11
just a curious question, has there ever been a series solution to the exact motion of undamped pendulum?
 

Related to Solving a Second-Order Differential Equation

1. What is a second-order differential equation?

A second-order differential equation is a mathematical equation that describes the relationship between a function and its second derivative. It is represented in the form of y'' + P(x)y' + Q(x)y = R(x), where y is the dependent variable, x is the independent variable, and y' and y'' represent the first and second derivatives of y, respectively.

2. What is the process for solving a second-order differential equation?

The process for solving a second-order differential equation involves finding a particular solution that satisfies the given equation. This can be done by using various methods, such as separation of variables, substitution, or the method of undetermined coefficients. Once a particular solution is found, it can be combined with the general solution of the homogeneous equation to obtain the final solution.

3. What is the difference between a homogeneous and non-homogeneous second-order differential equation?

A homogeneous second-order differential equation is one in which all the terms contain the dependent variable and its derivatives. On the other hand, a non-homogeneous second-order differential equation has additional terms that do not contain the dependent variable or its derivatives. The solutions to homogeneous equations are typically simpler, while non-homogeneous equations require more complex techniques to solve.

4. How do initial conditions affect the solution of a second-order differential equation?

Initial conditions, such as the values of y and y' at a specific point, are necessary to obtain a unique solution for a second-order differential equation. These conditions are used to determine the arbitrary constants in the general solution, making it a particular solution that satisfies the given equation.

5. What are some real-world applications of second-order differential equations?

Second-order differential equations are widely used in various fields, including physics, engineering, and economics, to model and analyze real-world phenomena. Some examples of their applications include predicting the motion of a spring-mass system, analyzing the behavior of electrical circuits, and determining the growth rate of populations in biology and economics.

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