A Trolley oscillates horizonally between two springs....

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Homework Help Overview

The discussion revolves around a trolley oscillating between two springs, focusing on the relationship between kinetic energy (K.E) and potential energy (P.E) at various positions. Participants explore how these energies compare, particularly at the equilibrium position and maximum amplitude.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants question how to determine the position where K.E equals P.E and how P.E varies with displacement. They discuss the implications of having two springs and the potential energy dynamics at the equilibrium position.

Discussion Status

The conversation is ongoing, with participants providing insights and questioning assumptions about energy relationships. Some guidance has been offered regarding the nature of potential energy in relation to kinetic energy, but no consensus has been reached on specific positions or values.

Contextual Notes

There is a focus on understanding the energy dynamics in a system with two springs, including assumptions about their states at equilibrium and the implications for potential and kinetic energy. Participants express confusion about how to relate these energies and their maxima.

  • #31
E/2 = 1/2kx^2
 
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  • #32
Mr Smailes said:
E/2 = 1/2kx^2
It looks as if you are trying to take two steps at once. How does this compare to what I learned, that ##E = {1\over 2} kx^2 ## ?
 
  • #33
BvU said:
It looks as if you are trying to take two steps at once. How does this compare to what I learned, that ##E = {1\over 2} kx^2 ## ?
I previously defined E in this thread as the max KE.
 
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  • #34
Mr Smailes said:
E/2 = 1/2kx^2
Right.
To be clear, let's call that displacement ##x_{E/2}##.
So how does that displacement compare with the maximum displacement, ##x_E##?
 
  • #35
E/2 = 1/2k(x/2)^2 and E = 1/2kx^2
 
  • #36
Mr Smailes said:
E/2 = 1/2k(x/2)^2 and E = 1/2kx^2

As a quick tip, it helps if you start learning how to post using LaTeX to post math equations on discussion forums like PF (see the LaTeX Guide link in the lower left of the Edit window). :smile:

$$ \frac{E}{2} = \frac{1}{2}k ({\frac{x}{2}})^2 $$
and ## E = \frac{1}{2} k x^2 ##

Did I parse your text correctly?
 
  • #37
Mr Smailes said:
E/2 = 1/2k(x/2)^2 and E = 1/2kx^2
That is clearly false. You seem to be assuming the answer, and a wrong one.

Please use the subscripted variables I defined, ##x_E, x_{E/2}## - or maybe define X as the max displacement and Y as the displacement when PE=KE - and write the above two equations using those.
 
  • #38
To be honest this has confused me further now, I am really not sure where I am going with these equations for a 1 mark question there must be a simpler way to visualise and understand this problem
 
  • #39
Mr Smailes said:
To be honest this has confused me further now, I am really not sure where I am going with these equations for a 1 mark question there must be a simpler way to visualise and understand this problem
Well, it really is not that difficult.
Write the equation relating the variables ##E, k, x_E##; write the equation relating the variables ##E, k, x_{E/2}##; solve to find ##x_{E/2}## in terms of ##x_E##.
 
  • #40
Can you show that?
 
  • #41
Mr Smailes said:
Can you show that?
I'm just asking for the equations you had in post #35 but with ##x_{E/2}## instead of "x/2" and ##x_E## instead of the other ##x##.
 
  • #42
Equate to XE/2=XE/2
 
  • #43
Mr Smailes said:
Equate to XE/2=XE/2
If you are saying that is the result you get by following what I asked you to do in post #41 then you have made a mistake. Please post your working.
 

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