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nas7232
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#1
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1)
A student was studying the motion of a simple pendulum the time period of which was given by T =2pie x (l/g)sqrooted

He measured T for values of l given by

l/m = 0/1 0/4 0/7 1.0 and plotted a graph of T against sqrRoot L in order to deduce a value of g, the free-fall acceleration.

Explain why these values of l are poorly chosen.


2) A skylab astronauts suggest that the calibration experient with 1kg mass could have been carried out on earth before take off. if a similiar experiment were conducted on earth would the time period be greater, less than, or equal to 1.22 s. Explain your answer.

[period of oscillation outside the earth of the mouse is 1.22 s]

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lukeD
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#2
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Question 1:
I think this might be to do with the fact that the square roots of 0.1, 0.4 and 0.7 are not very easy values to plot on a graph.
Also, four results might not be enough for an accurate reading.
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nas7232
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Cheers

I add the accuracy bit. All i put before is that if one/two values are outliers then the estimate of g will be serverly affected. [1 mark]
[physics homework_due in monday]
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nas7232
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Can anyone help me with Q2 please, or even guess. It's due in tomorow.

I may go for;

The period doesn't change because of g does not feature in the equation. Therefore, g has no effect.

But the question is for 3 marks.
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habosh
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for two it's the same,because mass is the same everywhere it doesn't change as the equation is t=(2*pi*)sqrt(m/k)
where k and 2pi are constants and the mass would be the same then the period is the same
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