Coupled pendulum-spring system

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Homework Help Overview

The discussion revolves around a coupled pendulum-spring system, focusing on the separation vector between two masses and its representation in terms of angular displacements. Participants are exploring the geometric and algebraic interpretations of the separation vector and its implications in the context of the problem.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants are questioning the necessity of introducing unit vectors for the masses and whether they can be simplified to a single unit vector for small perturbations. There is also confusion regarding the subtraction of arc lengths and the implications of squaring in the context of displacement.

Discussion Status

The discussion is actively exploring different interpretations of the separation vector and its components. Some participants are providing insights into the relationship between arc lengths and displacements, suggesting that for small angles, the distinctions may not be significant. However, there is no explicit consensus on the best approach or understanding of the problem.

Contextual Notes

Participants are grappling with the definitions and assumptions related to the displacement of the spring from its equilibrium position, particularly in the context of small angular displacements and their effects on the separation vector.

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Homework Statement
Please see below. I have a doubt about the solution of the problem.
Relevant Equations
Arc length = ##R\theta##
The problem and solution are,
1712980472532.png

1712980490956.png

1712980508579.png

However, I am confused how the separation vector between the two masses is

##\vec x = x \hat{k} = x_2 \hat{x_2} - x_1 \hat{x_1}= l\theta_2 \hat{x_2} - l\theta_1 \hat{x_1 } = l(\theta_2 - \theta_1) \hat{k}##. where I define the unit vector from mass 2 to mass 1 as ##\hat{k}## along the spring. Does someone please know of a algebraic or geometric proof for this?

Thanks a lot!
 
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ChiralSuperfields said:
Homework Statement: Please see below. I have a doubt about the solution of the problem.
Relevant Equations: Arc length = ##R\theta##

The problem and solution are,
View attachment 343276
View attachment 343277
View attachment 343278
However, I am confused how the separation vector between the two masses is

##\vec x = x \hat{k} = x_2 \hat{x_2} - x_1 \hat{x_1}= l\theta_2 \hat{x_2} - l\theta_1 \hat{x_1 } = l(\theta_2 - \theta_1) \hat{k}##. where I define the unit vector from mass 2 to mass 1 as ##\hat{k}## along the spring. Does someone please know of a algebraic or geometric proof for this?

Thanks a lot!
Which particular step worries you?
It seems a bit verbose to me. Why introduce ##\hat{x_2} ## and ## \hat{x_1}##? Aren’t they obviously ##\hat k## for small perturbations (compared to the length of the spring)?
 
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haruspex said:
Which particular step worries you?
It seems a bit verbose to me. Why introduce ##\hat{x_2} ## and ## \hat{x_1}##? Aren’t they obviously ##\hat k## for small perturbations (compared to the length of the spring)?
Thank you for your reply @haruspex!

Sorry are you saying that ##\hat{x_2} = \hat{x_1} = \hat{k}## for small ##\theta##? My confusion is how the spring displacement from equilibrium position is ##l\theta_2 - l\theta_1 = l\theta_1 - l\theta_2## since the displacement is squared. I am confused when I try to imagine subtracting two arc lengths from each other.

Thanks!
 
ChiralSuperfields said:
My confusion is how the spring displacement from equilibrium position is ##l\theta_2 - l\theta_1 = l\theta_1 - l\theta_2## since the displacement is squared. I am confused when I try to imagine subtracting two arc lengths from each other.

Thanks!
For small perturbations, the arc length is near enough the same as the displacement in the initial tangential direction, i.e. the ##\hat k## direction.
Since it is squared, it does not matter which way around you do the subtraction.
 
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