Is my graph of Ep/J against t/s for an oscillating object correct?

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SUMMARY

The discussion centers on the graph of potential energy (Ep) against time (t/s) for an oscillating object undergoing simple harmonic motion. The equation Ep = mgA sin(wt) is established, where m is mass, g is gravitational pull, A is amplitude, and wt is angular frequency. The participants confirm that while the shape of the graph is sinusoidal, the amplitude must be adjusted based on the reference point for potential energy. The final consensus is that the graph should reflect values above the x-axis, starting from zero, and not dip below it.

PREREQUISITES
  • Understanding of simple harmonic motion (SHM)
  • Familiarity with potential energy equations (Ep = mgh)
  • Knowledge of sinusoidal functions and their graphical representation
  • Basic principles of oscillatory motion
NEXT STEPS
  • Study the derivation of potential energy in oscillatory systems
  • Learn about the characteristics of simple harmonic motion graphs
  • Explore the impact of amplitude on potential energy in oscillating systems
  • Investigate the relationship between kinetic and potential energy in SHM
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and oscillatory motion, as well as educators seeking to clarify concepts related to potential energy graphs in simple harmonic motion.

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Homework Statement


Picture004.jpg

Picture005.jpg

Hi,the question is draw a graph of E_{p} against t/s of the oscillating motion above,I've done it and have no exact answer in my hand,hoping some ppl to double check my answer that's all.My answer is here below

Homework Equations


E_{p}=mgh , where E_{p} is potential energy,m=mass,g=gravitational pull,h=height


The Attempt at a Solution


q.jpg

Correct?!
 
Last edited:
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You see, Ep=mgx, where x is the height of the end of the strip. Now, since the motion is simple harmonic (from the graph), x=A sin(wt) (equation of shm), therefore, Ep=mgA sin(wt), which can be rewritten as Ep=B sin(wt), where B is a constant (mgA). From this it seems that your solution is incorrect... The graph will look like fig 4.3, but the amplitude will be different...
 
Well I've drawn my amplitude all above x-axis as the E_{p} starts from 0-origin,then increases to max then back to zero again,forming a sinusoidal wave type pattern.The graph requested didn't put a requirement to be exact in amplitude.How different u mean?
 
Different as in, taking the potential when the rod is at the mean position to be zero, the graph should be above and below the curve.
 
your answer is correct but what people are saying that u are missing the values on the y axis, they are not the same values as on the first graph,
but the shape of the curve is correct,

because the starting point is 0, no it does not go above and below the x axis
 
Hey thanks a bunch!Solved!**
 

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