# Compound Pendulum QuestionWatch

#1
I'm not sure if I am approaching this question correctly.

I have been asked to plot T^2 against h for a compound pendulum. The graph I obtained has the following shape if I remember correctly:

From the graph:
1) Determine the minimum time period
2) Determine the minimum distance
3) Estimate the uncertaintity in the minimum distance.

My thoughts are:
1) Draw a tangent to the graph at the very minimum point at T. Square root the value for the time period.
2) Observe the middle value of h that it could possibly be
3) Calculate the range of values it could be and half it.

Is this correct?
0
2 years ago
#2
(Original post by JimmyMcgill)
I'm not sure if I am approaching this question correctly.

I have been asked to plot T^2 against h for a compound pendulum. The graph I obtained has the following shape if I remember correctly:

From the graph:
1) Determine the minimum time period
2) Determine the minimum distance
3) Estimate the uncertaintity in the minimum distance.

My thoughts are:
1) Draw a tangent to the graph at the very minimum point at T. Square root the value for the time period.
2) Observe the middle value of h that it could possibly be
3) Calculate the range of values it could be and half it.

Is this correct?
My guess is that a marker would want to see 2 tangents drawn on slightly apart to show the uncertainty in finding the minimum. You want them both to be as flast as possible but one to curve down \ and one to point up a little /, Take the midpoint of these as the minimum and use half the range to estimate uncertainty.

Depends on how hard it is to get a tangent though. If you use a mirror, it ought to be possible to find the minimum within a very small uncertainty so you'd probably want to use the precision of your plotted point instead (?? Not sure about this, clearly a graph of loads of points can in fact have lower uncertainty than each point. ??)
1
#3
(Original post by lerjj)
My guess is that a marker would want to see 2 tangents drawn on slightly apart to show the uncertainty in finding the minimum. You want them both to be as flast as possible but one to curve down \ and one to point up a little /, Take the midpoint of these as the minimum and use half the range to estimate uncertainty.

Depends on how hard it is to get a tangent though. If you use a mirror, it ought to be possible to find the minimum within a very small uncertainty so you'd probably want to use the precision of your plotted point instead (?? Not sure about this, clearly a graph of loads of points can in fact have lower uncertainty than each point. ??)
Can you provide an example with the tangents? I'm not sure what you mean.
0
2 years ago
#4
(Original post by JimmyMcgill)
Can you provide an example with the tangents? I'm not sure what you mean.

I'm not feeling this as much though now tbh. I guess it's up to you to decide whether the uncertainty in finding your minimum value is to do with the uncertainty drawing the tangents or the uncertainty associated with your line of best fit.

This gives you a way of getting the uncertainty in finding the minimum of your curve, but if you think that most of the uncertainty lies in your curve then you've got to do something else.

What level is this question aimed at?
0
#5
(Original post by lerjj)

I'm not feeling this as much though now tbh. I guess it's up to you to decide whether the uncertainty in finding your minimum value is to do with the uncertainty drawing the tangents or the uncertainty associated with your line of best fit.

This gives you a way of getting the uncertainty in finding the minimum of your curve, but if you think that most of the uncertainty lies in your curve then you've got to do something else.

What level is this question aimed at?
A-level. I can't see it being the uncertainty of the line of best fit because it's not on the spec to my knowledge.

But I'm thinking the tangents is also too complicated. It might simply be the uncertaintity in reading from the scale and is therefore half the precision of the graph graduation.
0
2 years ago
#6
(Original post by JimmyMcgill)
A-level. I can't see it being the uncertainty of the line of best fit because it's not on the spec to my knowledge.

But I'm thinking the tangents is also too complicated. It might simply be the uncertaintity in reading from the scale and is therefore half the precision of the graph graduation.
TBH I think I'd just take whatever the smallest increment on your graph is. If you had a really wide minimum then it might be hard to find the absoulte bottom by eye, so finding two tangents that lean \ and / that are near the bottom and averaging them gives you a decent guess. But equally, you could just use a mirror and get it almost dead on.

If it's A level then they can't reasonably expect you to do any sort of calculation involving the line of best fit I don't think. What you'd normally do is plot error bars and then draw two curves that both go through the error bars, but have different minima positions, and use those as your range. That's a lot of effort without a computer, and I'm pretty sure not on an A level spec (plus there's probably some better statsy way to do it anyway).
0
2 years ago
#7
I could be wrong, but shouldn't the graph of T^2 against l be a straight line for a pendulum?

Part 1 seems simple enough, draw a tangent at the minimum and root it.

I don't understand what you mean by part 2. The minimum distance (h?) is surely the value of h closest to the y axis?

Part 3 I imagine is either your answer to part two +/- either the graph scale or the graph scale plus precision of instrument.
0
#8
(Original post by Einstein1997)
I could be wrong, but shouldn't the graph of T^2 against l be a straight line for a pendulum?

Part 1 seems simple enough, draw a tangent at the minimum and root it.

I don't understand what you mean by part 2. The minimum distance (h?) is surely the value of h closest to the y axis?

Part 3 I imagine is either your answer to part two +/- either the graph scale or the graph scale plus precision of instrument.
This is a compound pendulum, a rod with the centre of mass in the middle.

Part 2 is the value of h which gives the minimum time period.
0
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