This'll be quick-For an oscillating object as shown,it's graph of Ep/J against t/s ?

[inv]

1. The problem statement, all variables and given/known data
http://img.photobucket.com/albums/v282/asdas/Picture004.jpg
http://img.photobucket.com/albums/v282/asdas/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

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


3. The attempt at a solution
http://img.photobucket.com/albums/v282/asdas/q.jpg
Correct?!

[chaoseverlasting]

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...

[inv]

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?

[chaoseverlasting]

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.

[esalihm]

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

[inv]

Hey thx a bunch!Solved!**