Differential equation for a pendulum

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Homework Help Overview

The discussion revolves around deriving the equation of motion for a simple pendulum, which involves a mass on a string and its angular displacement from the vertical. The original poster seeks clarification on the correct formulation of the motion equation and the conditions under which it can be approximated as simple harmonic motion (SHM).

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to derive the equation of motion and expresses confusion regarding the interpretation of the angle and the general solution of a specific differential equation. Some participants question the clarity of the angle definition and suggest that a figure would aid understanding. Others discuss the implications of additional terms in differential equations and their effect on solutions.

Discussion Status

The discussion is ongoing, with participants providing insights and suggestions for clarification. Some guidance has been offered regarding the simplification of the sine function and the structure of the differential equation. However, there is no explicit consensus on the original poster's understanding of the problem.

Contextual Notes

There are indications of potential misunderstandings regarding the formulation of the equation of motion and the nature of the differential equations involved. The original poster is also navigating the implications of constants in the equations, which may not have been fully addressed.

Faiq
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Homework Statement


A simple pendulum is formed by a light string of length ##l## and with a small bob ##B## of mass ##m## at one end. The strings hang from a fixed point at another end. The string makes an angle ##\theta## with the vertical at time ##t##. Write down an equation of motion of ##B## perpendicular to ##OB## and state the approximation that enables you to treat this equation as an SHM equation. State the general solution of this equation.

The Attempt at a Solution


Equation of motion : ## \rm \large -mgl\sin(\pi -\theta) = ml^2 \ddot \theta##
Rearranges to: ## \rm \large \ddot \theta =-\frac{g}{l}sin(\pi-\theta)##
Approximation: ## \rm \large sin \theta \approx \theta## provided ## \theta ## is small.
General solution: ## \rm \large \theta = \theta_0sin(\sqrt{\frac{g}{l}}t)##

Are my answers correct?
The reason I am confused is because I think I have misunderstood what they mean by "equation of motion of ##B## perpendicular to ##OB##" and may have got all of my answers wrong.
And secondly I am well aware of finding a general solution of a differential equation of the form ## \rm \small \ddot y =ay ##, however, I have never seen a general solution of a differential equation of the form ##
\rm \small \ddot y =a(\lambda-y) ## where ##\lambda## is a constant.
 
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It is not entirely clear what you are calling theta (and thus why you have involved a pi - theta argument for the sine). A figure would really help your question, along with labels on the figure.

As to the solution for the differential equation, did you really mean ddy = a*y, where a is presumed to be positive? Or did you intend to have it be negative?

The additional term a*lambda simply adds a constant to the particular solution.
 
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Why are you interested in the solution to that differential equation?
 
Dr.D said:
It is not entirely clear what you are calling theta (and thus why you have involved a pi - theta argument for the sine). A figure would really help your question, along with labels on the figure.
WhatsApp Image 2017-04-13 at 5.22.17 AM.jpeg

I hope this will work.

Dr.D said:
As to the solution for the differential equation, did you really mean ddy = a*y, where a is presumed to be positive? Or did you intend to have it be negative?

The additional term a*lambda simply adds a constant to the particular solution.
Yes, can you show me how to add the constant to the particular solution? Like for example if the solution to the differential equation ## \ddot y = ay## where a is a constant is ## y = y_0 \sin \omega t##, how will the constant term be added in the solution if the equation is changed to ## \ddot y = a(\lambda-y)##? (If changes vary from equation to equation, please refer to the SHM equation given in my first post) If you find my example vague, feel free to make one of yours.
Chestermiller said:
Why are you interested in the solution to that differential equation?
I am not interested in that particular equation. I am just concerned whether changes will occur in the solution if the concerned variable was increased or decreased by a constant term.
 
Start by replacing sin(pi-theta) with just sin(theta); they are exactly equal.

Then, with a little bit of re-arrangement, your equation should look like
dd(theta) + w^2*sin(theta) = 0

Then make the approximation that for small theta, sin(theta) = theta, so you have
dd(theta) + w^2*theta = 0

That should be the only equation you need to solve for this problem.
I really don't know where the question about a constant term on the right came from.
 
Okay thank you that explains a lot.
 

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