# I am not sure the source of error of simple pendulum experiment

Raul Lai
Aim: To measure the acceleration due to gravity using a simple pendulum.

source of error

1. The length of thread between the bob and the coins is not measured by very accurate instruments. Since meter ruler has a large percentage error, as a result, it should be replaced by a vernier caliper.
2. The reaction time of human may also introduce an error to the measurement of 20 completed oscillations of the bob. It can be reduced by using rhythm to count before started.
3. The damping occurred to reduce the amplitude of SHM. The time for counting the oscillation is shorted. The driving frequency of the external force should be closer to the natural frequency of the system and the angle between the bob and vertical should be small in order to diminish the damping.
4. The bob started rotating by the impact of the Earth’s revolution. It disturbs the measurement of counting the oscillation. The experiment should start quickly after moving the bob.

Is that correct?

And one more thing, do the period against length (the string) graph pass through the origin?
The result in my experiment didn't.

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betel
Welcome to the forum.
Maybe you should first describe your experiment.
What is your setup with coins and bob?
For a usual simple pendulum I do not see where your external force should come into play.

Raul Lai
Oh I see, my set up seems like: http://www.practicalphysics.org/go/Experiment_480.html [Broken]
and move the weight (bob) to a height so that the attached thread is taut and makes an angle of 10° with the vertical.
counting the time for 20 completed oscillation.
draw a period square against length graph to find the slope and then the value of gravity

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Staff Emeritus
What did you expect to see, and what did you see? What are the relevant equations here?

Raul Lai
I want to expect to see the graph is pass through the origin as T=2π* root(L/g),
My result is here
Pendulum length/ m 1.5 1.25 1 0.75 0.5
Average t 49.71 45.315 40.58 35.34 29.20
(Period)2 T2/s2 6.179 5.134 4.117 3.122 2.132

But the graph didn't pass the origin.

betel
Well that is the difference between theory and experiment. Have you calculated the value for g you get?

Then you can try to find reasons for why you are getting an offset. And try thinking a bit yourself and do not blindly copy answers from somewhere else.

Raul Lai
no, all my classmates did the same result, I don't know why and I can't explain it.

betel
All got exactly the same result for the slope and the offset?

The answers you posted in your original post obviously come from some other similar experiment and you copied them before thinking if they are relevant.

Raul Lai
No, I mean my classmates plot the graph are also cannot pass through the origin.
One important thing is I cannot copy because this is for public exam.
The gravity I find is about 9.755.
And I just want to know the explanation.

betel
Well if your classmates use the same data there is no wonder they get the same result as you do.
So did you really do the experiment or are you just given the data?

Let's start with the possible explanations you gave in your first post. Could one of them be a solution? Which are relevant at all?

Raul Lai
Is that due to the natural frequency and hence the period exist when the length is zero.

betel
What is natural frequency? Mind: Your plot would indicate a negative frequency at zero length. Seems a bit unrealistic.

Did you calculate or plot the variance of your data?

Raul Lai
when length is 05.m, period square is 2.132
when length is 1.5m, period square is 6.179
slope is 4.047 positive

betel
Hm. you have five data points and use 2 to calculate your slope? You either have to calculate it using all points (e.g. excel can do that) or read of the slope from the graph you plotted. Just taking 2 points is not enough.
The reason why you measure for so many points is to get some redundancy and statistics. Maybe your points are spread so far that there is not "unique" way to draw the line. Then an offset will not be a problem.

And the answer to my last post?

Raul Lai
According to my textbook, it says
a simple pendulum will oscillate at a frequency known as its natural frequency (f).
This depends on it length.
equation: f=1/(2π)root(g/L) but now L is zero.....

betel
So it will not oscillate at all. What value of period do you read of for L=0 from your plot?

Raul Lai

betel
That is a bit too large, but you will still get a result that is to big including your confidence interval. Then we have to search for some systematics, what went wrong during your "experiment".

Raul Lai
I got another problem, when I put g=9.81 into the equation to find period square and plot a graph. It still cannot pass the origin.

betel
What do you mean? If you give yourself the length and g you can plot a perfect graph through the origin. It will not lie on your data points though.

betel
From the looks of it, whoever gave you the data tweaked it, such that it shows a systematic error. The statistics on the data points is too good.

Raul Lai
systematic error? do you mean the length of my meter ruler is wrong mark. But this is bizarre, it is too large.

Raul Lai
maybe air resistance?

betel
What impact does a wrong ruler have? What does air resistance do? Try to answer these questions first.

Raul Lai
if the ruler reading switched, all the length measured are also switched. By the equation, T square=4πsqure(L/g), if L get greater, the graph will switch upward.
if the air resistance exist, the period will get larger, by T=2πroot(L/g),the graph will also switch upward.

betel
Correct. But a shift upwards is not what we need. So air resistance should not play a role. A wrong ruler could still cause the error.
Any other possibilities?

Raul Lai
I don't think the the time for oscillation get systematic error, I count it by a stopwatch.

betel
Maybe your stopwatch has an error. Or you press the stop button always too late.

Raul Lai
I count it with rhythm...3,2,1,0,1,2,3...the reaction time delay is deminished

Raul Lai
It is 11pm in HK, I have no time

betel
Well, the counting part will not improve accuracy here that much. Much more important for the time accuracy is measuring 20 cycles, which will greatly reduce the error.
Well, the ruler seems so far the best possible answer. There could still be others.

Raul Lai
but it is impossible, as other group did the result of slope is 4.0875
it similar to me. beside, the error is too large

betel
Did you do the experiment or did somebody just give you the data?
When you fit a linear interpolation to your data you always need two parameters: slope and intersect. You will always get errors on both of them. But in this case, for your data, the statistical error cannot account for your wrong intersect. The intersect will be nonzero with 99.7% confidence.
So you need some explanation for it.

Raul Lai
the experiment I did is by my own hands
but, it is fine. I give up now. I should change the data to fit it
Thx for your help, it wasted you so much time
anyway, thank you.

betel
Never change data to fit theory! That is the way to hell for a physicist.
If measure something different you have to stand to your results. Propose possible error sources and possible solutions.
But never ever cheat on your data!