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    In the June 2010 AQA 6X paper, Section 2 Task B, the question asks you to sketch how you positioned your fiducial mark to investigate the time period for an oscillating chain of paper clips. It then asks you to explain why. The first mark point was obviously for placing it at the equilibrium, but the second mark given in the mark scheme was that the fiducial mark should be placed 'below some 3/4 of the length of the chain, and ideally positioned below the end of the chain'. In the examiners report, they commented that 'a worrying number of students positioned the fiducial mark behind the chain, and not below. relatively few could explain why the fiducial mark should be positioned like so'. My physics teacher does not understand why it should be ideally beneath the chain, or why it is not acceptable to place it behind. Can anyone explain the reasons behind this?
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    I think I can:
    Speed is max at the centre of the oscillation, yes? So, the chain will be moving at max velocity when it is at the bottom of the swing, therefore the uncertainty of your reading will go down compared to if you were to measure it at the edge of the oscillation where it is slowest. Placing the fiducial marker behind and below the chain ensures that you will be able to see it as the chain passes.
    I don't know why they only wanted it below the chain in the mark scheme. *sigh* AQA examiners…
    I hope I helped clear that up a little.
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    (Original post by Pirateprincess)
    I think I can:
    Speed is max at the centre of the oscillation, yes? So, the chain will be moving at max velocity when it is at the bottom of the swing, therefore the uncertainty of your reading will go down compared to if you were to measure it at the edge of the oscillation where it is slowest. Placing the fiducial marker behind and below the chain ensures that you will be able to see it as the chain passes.
    I don't know why they only wanted it below the chain in the mark scheme. *sigh* AQA examiners…
    I hope I helped clear that up a little.
    Why is the uncertainty less when the speed is greatest at the centre of oscillation?
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    (Original post by rainerised)
    Why is the uncertainty less when the speed is greatest at the centre of oscillation?
    because when it is moving slowly it is harder to accurately judge the maximum point, whereas it is easier to record accurately the point at which it passes the fiducial mark when it is moving quickly. It is still affected by random errors in timing (human errors) and so use of a data logger and light gates might be applicable.
 
 
 
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