A Trolley oscillates horizonally between two springs....

[berkeman]

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).

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

Did I parse your text correctly?

 

[haruspex]

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.

 

[Mr Smailes]

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

 

[haruspex]

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 $x_E$.

 

[Mr Smailes]

Can you show that?

 

[haruspex]

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 $x$.

 

[Mr Smailes]

Equate to X_E/2=X_E/2

 

[haruspex]

Equate to X_E/2=X_E/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.