Finding Linear Graph for Pendulum Dilemma Equation

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Homework Help Overview

The discussion revolves around finding a linear graph representation for the equation T=2π(√L/√g), which relates the period of a pendulum to its length and the acceleration due to gravity. Participants are exploring how to appropriately graph this relationship in a linear format.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the appropriate variables to plot on the axes, with suggestions to plot T² on the y-axis and L on the x-axis. There is also a question regarding the interpretation of slope in the context of the graph.

Discussion Status

Some participants have confirmed the proposed graphing method, while others are questioning the definition of slope in this context. There is an ongoing exploration of the relationship between the variables and the implications for the slope of the graph.

Contextual Notes

Participants are navigating through the definitions and relationships inherent in the equation, with some confusion regarding the slope and its calculation. The discussion reflects a mix of understanding and uncertainty about the graphing process.

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Homework Statement



I am suppose to find a linear graph for the equation T=2pie(√L/√g)

Homework Equations





The Attempt at a Solution



The best linear graph I could think of was L/T^2.

Am I doing it right? Thanks.
 
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Yes.

T[itex]^{2}[/itex] on y-axis and L on x-axis.
 
grzz said:
Yes.

T[itex]^{2}[/itex] on y-axis and L on x-axis.

Thank you very much. This has been like a huge thorn in my finger all day!
 
I thought the slope was always y/x, so why isn't the slope of the graph T^2/L?
 
Slope is NOT y/x but slope = [itex]\Delta[/itex]y/([itex]\Delta[/itex]x)
or more exactly slope = dy/(dx).
 
T = 2∏√(L/g)

therefore T[itex]^{2}[/itex] = 4π[itex]^{2}[/itex]L/g

so d(T[itex]^{2}[/itex])/dL = 4π[itex]^{2}[/itex]/g = slope
 
grzz said:
T = 2∏√(L/g)

therefore T[itex]^{2}[/itex] = 4π[itex]^{2}[/itex]L/g

so d(T[itex]^{2}[/itex])/dL = 4π[itex]^{2}[/itex]/g = slope

oh that makes much more sense. thank you.
 

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