Measuring 'g' and systematic errors

[Jimmy87]

Homework Statement

Figure 2.2 (attached) shows a graph of distance against time squared for a ball bearing falling through the air. In accordance with s = ut + 1/2 a t^2 the graph should be a straight line through the origin. Explain how a systematic error in s and t could lead to the graph not going through the origin and state what effect, if any, each would have on the gradient.

Homework Equations

s = ut + 1/2 a t^2

The Attempt at a Solution

So I have said if 's' has the systematic error then the plotted values for 's' are too small and if the error was in 't' then the plotted values are too big. The gradient wouldn't change if 's' has the error as they would all shift by the same amount. If the error was in 't' then the gradient must change because t^2 is plotted so a constant error in t would lead to an increasing error in t^2. I said the gradient would get shallower but the answer says it should get steeper and I don't get why (unless the answer is wrong). If the error in t^2 gets bigger (which the answer also says) then each plot on the graph would shift to the right but by an increasing amount as you consider higher values so surely this is a shallower gradient? For example, let's say the first plotted value shifts right by 1 small square, the second by 2 small squares and so on then the gradient decreases?

 

[kuruman]

So I have said if 's' has the systematic error then the plotted values for 's' are too small and if the error was in 't' then the plotted values are too big.

What makes you say this? Why would a systematic error in s give plotted values that are too small? Why not too large? First you need to explain what (in terms of what you did in the lab) would lead to a systematic error in s and in t and in what direction that systematic error would drive the values of s and t, larger or smaller? Also, it seems that you have an initial velocity u. Did you measure it? If so, can a systematic error be attached to it as well? Related to the u issue is that your plot shows s (dependent variable in m) vs. t² (independent variable in s²). Why do you expect a straight line if your equation is
##s = ut + 1/2 a t^2## ? If I let ##x = t^2##, then ##s = u \sqrt{x}+\frac{1}{2} a x.## This is not the equation of a straight line.

 

[haruspex]

The gradient wouldn't change if 's' has the error

Right.

If the error was in 't' then the gradient must change because t^2 is plotted so a constant error in t would lead to an increasing error in t^2. I said the gradient would get shallower

Doesn't it depend which way the error is?
Suppose the error is Δt. What slope would you get?

you have an initial velocity u

what you did in the lab

As I read it, the ball is released from rest. I think Jimmy included the u as part of the standard equation. (Jimmy, it will be clearer if you delete u from your equations.)
Also, the lab may be fictitious.

 

[kuruman]

In accordance with s = ut + 1/2 a t^2 the graph should be a straight line through the origin.

Which graph? There is no mention of what the dependent and independent variables are supposed to be. The usual way of analyzing constant acceleration measurements in the laboratory is to plot s/t vs. t. This linearizes the graph so that twice the slope is the acceleration and the intercept is the initial velocity. Did @Jimmy87 misunderstand what to plot and how?

Even if the object were supposed to be released from rest, the hard truth is that it cannot be done reproducibly so that ##u = 0 \pm~ 0 ~m/s##. If the lab is fictitious, then one still has to assume a lab procedure for conducting the experiment and imagine what the sources of systematic error can possibly be.

 

[Jimmy87]

Right.

Doesn't it depend which way the error is?
Suppose the error is Δt. What slope would you get?As I read it, the ball is released from rest. I think Jimmy included the u as part of the standard equation. (Jimmy, it will be clearer if you delete u from your equations.)
Also, the lab may be fictitious.

Sorry guys, I should have made things clearer. This is not an experiment we did, this is an exam style question. I have attached the background info to the question to make it clearer. The exact question is "Explain how a systematic error in s and t could lead to the graph not going through the origin and state what effect, if any, each would have on the gradient." The way the answer is worded tells me this question is basically saying the graph should go through the origin but it doesn't (due to systematic error) so what would have to happen to s and t for it to do this and how these changes would affect the gradient. The bit I got wrong is it says "a constant systematic error in t would lead to an increasing error in t^2 so the gradient would be steeper" whereas I said shallower. I still don't get how an increasing error in t would make the points steeper?

 

[Jimmy87]

Right.

Doesn't it depend which way the error is?
Suppose the error is Δt. What slope would you get?As I read it, the ball is released from rest. I think Jimmy included the u as part of the standard equation. (Jimmy, it will be clearer if you delete u from your equations.)
Also, the lab may be fictitious.

Just saw your point about which way the error was. Yes! That is my point! Surely it depends which way the error is so how can the answer give an increasing gradient?

 

[haruspex]

Surely it depends which way the error is

Right, but let us say the systematic error is positive. Would that make the observed gradient more or less than it should be?

 

[Jimmy87]

Right, but let us say the systematic error is positive. Would that make the observed gradient mpre or less than it should be?

Would than not make the gradient less? If we say there is no systematic error and the line goes through the origin. If we now add a positive error in t then we would get an increasing error in t^2. Since t^2 is on the x-axis would each point not shift right by an increasing amount?

 

[haruspex]

Would than not make the gradient less? If we say there is no systematic error and the line goes through the origin. If we now add a positive error in t then we would get an increasing error in t^2. Since t^2 is on the x-axis would each point not shift right by an increasing amount?

Right.
Sorry, I forgot you already wrote that in post #1.

 

[kuruman]

Sorry guys, I should have made things clearer.

Thank you for making things clearer. I now understand the scope and context of your post. Perhaps you could see better what might be going on if you first assume systematic errors ##\Delta s## and ##\Delta t##, which could be positive of negative. Then consider an "experimental" expression ##f_{expt.} = (s + \Delta s) -[ \frac{1}{2}a(t+ \Delta t)^2]## and a "theoretical" expression ##f_{theo.} = s - \frac{1}{2}at^2##. Find the difference ##\Delta f##; it should indicate, point by point, which way the systematic errors would shift the "experimental" line relative to the "theoretical" line. Note: you expect ##\Delta t << t## so you can drop the ##(\Delta t)^2## term in the expansion.

 

[Jimmy87]

Right.
Sorry, I forgot you already wrote that in post #1.

So is the answer wrong? Unless the question meant to ask about the true gradient which would be steeper?

 

[haruspex]

So is the answer wrong?

I'd say so.

Unless the question meant to ask about the true gradient

Not sure what you mean. Can you explain more?