- #31
Mr Smailes
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E/2 = 1/2kx^2
A Trolley oscillates horizonally between two springs....
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SUMMARY
The discussion centers on understanding the relationship between kinetic energy (K.E.) and potential energy (P.E.) in a system involving a trolley oscillating between two springs. Participants clarify that K.E. and P.E. are equal at a specific displacement between the equilibrium position and maximum amplitude. The maximum potential energy occurs at the extremes of the oscillation, while kinetic energy is maximized at the equilibrium position. The conversation emphasizes the conservation of energy principle, stating that the sum of K.E. and P.E. remains constant throughout the motion.
PREREQUISITES- Understanding of Kinetic Energy (K.E.) and Potential Energy (P.E.) concepts
- Familiarity with Hooke's Law and spring constants
- Basic knowledge of oscillatory motion and equilibrium positions
- Ability to manipulate and solve equations involving energy conservation
- Study the principles of conservation of mechanical energy in oscillating systems
- Learn how to derive equations for K.E. and P.E. in spring systems
- Explore the effects of varying spring constants on oscillation behavior
- Investigate real-world applications of oscillatory motion in engineering and physics
Students studying physics, particularly those focusing on mechanics and oscillatory systems, as well as educators seeking to clarify concepts related to energy conservation in spring systems.
- #32
BvU
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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 ## ?Mr Smailes said:
- #33
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I previously defined E in this thread as the max KE.BvU said:
- #34
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Right.Mr Smailes said:
To be clear, let's call that displacement ##x_{E/2}##.
So how does that displacement compare with the maximum displacement, ##x_E##?
- #35
Mr Smailes
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E/2 = 1/2k(x/2)^2 and E = 1/2kx^2
- #36
berkeman
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Mr Smailes said:
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?
- #37
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That is clearly false. You seem to be assuming the answer, and a wrong one.Mr Smailes said:
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
Mr Smailes
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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
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Well, it really is not that difficult.Mr Smailes said:
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
Mr Smailes
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Can you show that?
- #41
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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 said:
- #42
Mr Smailes
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Equate to XE/2=XE/2
- #43
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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.Mr Smailes said:
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