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1. Simple Harmonic Motion and energy of the system question
http://www.xtremepapers.com/CIE/Inte...2_w07_qp_4.pdf

Could anyone have a look at question 3 b (ii) 2 ?

The examiner's report says

"(ii) Many candidates who attempted this part of the question did state a frequency of 4 Hz. However, a fractional reduction in frequency equal to the fractional reduction in total energy was a common misconception. Candidates did not appear to realise that the graph would have the same basic shape, regardless of total energy. Thus, reading the value of x at (2.56 – 1.00) mJ would give the amplitude for a total energy of 1.00 mJ."

I don't understand this point.. Why would reading the amplitude value of a curve at KE= 1.56 give the correct answer? ANy help is much appreciated
Last edited by bmqib; 03-05-2012 at 21:50.
2. Re: Simple Harmonic Motion and energy of the system question
The answer is x should be around 0.5 cm, but assuming it decreases proportionately with decrease in energy, I get around 0.3 cm, I can try to post the graph if it would help
3. Re: Simple Harmonic Motion and energy of the system question
(Original post by bmqib)
http://www.xtremepapers.com/CIE/Inte...2_w07_qp_4.pdf

Could anyone have a look at question 3 b (ii) 2 ?

The examiner's report says

"(ii) Many candidates who attempted this part of the question did state a frequency of 4 Hz. However, a fractional reduction in frequency equal to the fractional reduction in total energy was a common misconception. Candidates did not appear to realise that the graph would have the same basic shape, regardless of total energy. Thus, reading the value of x at (2.56 – 1.00) mJ would give the amplitude for a total energy of 1.00 mJ."

I don't understand this point.. Why would reading the amplitude value of a curve at KE= 1.56 give the correct answer? ANy help is much appreciated

It's not obvious, I agree.
The PE curve is the inverse of the KE curve. (The sum of the two, total energy, is a constant straight line at height 2.56mJ).

What would the width of the PE curve be at 1mJ?
It would represent the amplitude at 1mJ
This, by symmetry, is the same value as the amplitude of the KE curve at 1.56mJ
4. Re: Simple Harmonic Motion and energy of the system question
(Original post by Stonebridge)
It's not obvious, I agree.
The PE curve is the inverse of the KE curve. (The sum of the two, total energy, is a constant straight line at height 2.56mJ).

What would the width of the PE curve be at 1mJ?
It would represent the amplitude at 1mJ
This, by symmetry, is the same value as the amplitude of the KE curve at 1.56mJ
OK, but why would the width of the PE curve be the amplitude? As the total energy drops to 1 mJ, the Max PE would have also dropped to 1 mJ. We can thus no longer consider the first curve, as the constant straight line drops to 1 mJ (but the shape of new curve would be similiar to the original). So Why can't we just take the new amp,itude of the lower curve? What's the mistake in my understanding? When we have lower total energy, shouldn't the curve be completely different?
5. Re: Simple Harmonic Motion and energy of the system question
If you can wait until tomorrow afternoon I will draw the curves for you. I don't have time just now.
Until then, think of the given PE curve for this motion, and draw a horizontal line across at 1mJ.
Now invert that smaller PE curve to get the new KE curve.
What is the amplitude of the new KE curve?
6. Re: Simple Harmonic Motion and energy of the system question
(Original post by bmqib)
OK, but why would the width of the PE curve be the amplitude? As the total energy drops to 1 mJ, the Max PE would have also dropped to 1 mJ. We can thus no longer consider the first curve, as the constant straight line drops to 1 mJ (but the shape of new curve would be similiar to the original). So Why can't we just take the new amp,itude of the lower curve? What's the mistake in my understanding? When we have lower total energy, shouldn't the curve be completely different?
If the total energy is 1mJ..... Then the max KE is 1mJ i.e. when it is at the equilibrium position.

So 0.5mv^2 = 1mJ

v^2 = w^2 (a^2 -x^2) x =0 and a = amplitude

Sub into the previous equation and solve for a.

7. Re: Simple Harmonic Motion and energy of the system question
8. Re: Simple Harmonic Motion and energy of the system question
(Original post by Stonebridge)
Thanks so much, swear half og A2 I've learnt from just reading your explanations here. just clarifying this confusion, isn't amplitude half of the width? I mean the range is from Xo to -Xo after all.
Last edited by bmqib; 04-05-2012 at 15:54.
9. Re: Simple Harmonic Motion and energy of the system question
(Original post by bmqib)
Thanks so much, swear half og A2 I've learnt from just reading your explanations here. just clarifying this confusion, isn't amplitude half of the width? I mean the range is from Xo to -Xo after all.
If you mean amplitude is equal to half length of DF.... The yes.
10. Re: Simple Harmonic Motion and energy of the system question
(Original post by bmqib)
Thanks so much, swear half og A2 I've learnt from just reading your explanations here. just clarifying this confusion, isn't amplitude half of the width? I mean the range is from Xo to -Xo after all.
Yes indeed. The amplitude is half of that distance. I was being a bit sloppy with my terms.