A Trolley oscillates horizonally between two springs....

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To determine where kinetic energy (K.E.) equals potential energy (P.E.) in a system with two springs, it's essential to understand that at the equilibrium position, both springs have zero P.E. As the trolley moves away from this position, the P.E. of each spring increases, while the K.E. decreases, maintaining the conservation of energy. When K.E. and P.E. are equal, both will be lower than their maximum values, occurring at a point between the equilibrium position and maximum amplitude. The maximum P.E. occurs at the extremes of displacement, where K.E. is zero. Understanding these relationships is crucial for solving problems related to oscillatory motion in spring systems.
  • #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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