Explaining Pendulum: Period & Length Proportionality

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Homework Help Overview

The discussion revolves around the relationship between the period of a pendulum and its length, specifically exploring why the period is proportional to the square root of the length. The subject area includes concepts from mechanics and oscillatory motion.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the physical parameters that might influence the period of a pendulum and how these parameters can be combined to derive a dimension of time. There are attempts to relate potential energy and speed to the period, with some expressing confusion about the physics involved.

Discussion Status

The discussion includes various attempts to clarify the relationship between the pendulum's period and its length. Some participants express uncertainty about the physics concepts, while others provide mathematical expressions related to the period. There is no explicit consensus, but the conversation is moving towards understanding the formula for the period.

Contextual Notes

Some participants indicate a lack of confidence in their understanding of physics, which may affect their engagement with the problem. The discussion includes references to specific formulas and approximations, suggesting a focus on mathematical relationships.

warriorofrovac
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Why is the period of the pendulum proportional to the square root of the length?
 
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Well, what physical parameters would you think might affect the period?
How can you combine these parameters in order to get the physical dimension of "time" (i.e, the dimension of the period)
 
The vertical displacement of a pendulum during its swing is approximately [itex]L \theta^2 / 2[/itex] (using the small angle approximation for the cosine). Multiply by m g to get for the available potential energy for its motion.

Also, the speed of the pendulum as it passes its lowest point will be about

[tex]v \sim \frac {L \theta}{T}[/tex]

Now set

[tex]g L \theta^2 / 2 \sim v^2/2 \sim \frac {1}{2} \left(\frac {L \theta}{T}\right)^2[/tex]

to find [itex]T \sim \sqrt {L/g}[/itex].
 
Last edited:
Tide said:
The vertical displacement of a pendulum during its swing is approximately [itex]L \theta^2 / 2[/itex] (using the small angle approximation for the cosine). Multiply by m g to get for the available potential energy for its motion.

Also, the speed of the pendulum as it passes its lowest point will be about

[tex]v \sim \frac {L \theta}{T}[/tex]

Now set

[tex]g L \theta^2 / 2 \sim v^2/2 \sim \frac {1}{2} \left(\frac {L \theta}{T}\right)^2[/tex]

to find [itex]T \sim \sqrt {L/g}[/itex].

erm..can i say..i have no idea what you just said..physics isn't my strong point :cry:
 
warriorofrovac said:
erm..can i say..i have no idea what you just said..physics isn't my strong point :cry:

Well, you did ask a physics question in a physics forum so ... ! :)
 
Okay, forget the why question

What if the simple formula showing that the period of the pendulum proportional to the square root of the length?
 
Okay,

[tex]T = 2 \pi \sqrt { \frac {L}{g}}[/tex]
 
Ah-ha. I thought so. Well i knew there was a 2pi in there somewhere..
I don't need to say why its like that...thankfully.
Hopefully that should be all the help i need...
 

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