Static measurement of Spring constant with error value

[RJLiberator]

Homework Statement

It's been a little while since I did one of these.
My question is: We have to use http://phet.colorado.edu/sims/mass-spring-lab/mass-spring-lab_en.html
to find the measurement of k, the spring factor using two different methods.

For the method I am looking at, is the static method.

Homework Equations

F(x)=-kx
F(x)=-k(x-x_0)
plot L(mass)=(x-x_0) = -F/k = -mg/k

The Attempt at a Solution

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What I am doing is using the various weights and measuring the displacement with the giving centimeter ruler.

As far as error measurement
I understand the each calculation of displacement will by +/- 0.5 cm as the error measurement.

My work:
So, if I have a table of values (.05kg, .05m), (.1kg, .1m), (.25kg, .24m)
I can use these values to calculate k, the spring constant. From the equation: K=-m*g/(x-x_0)

From point one, we see K= - (0.05kg*9.8m/s^2 )/ (0.05m) = -9.8N/m
From point two we see K = - (.1kg*9.8m/s^2)/(.1m) = -9.8 N/m
From point three we see K = - (.25kg*9.8m/s^2) / (.24m) = -10.21 N/m

conclusion
Here I have found the value of K.
Did I find it correctly?
Now for error the question states: "For example, how accurately can you measure the displacement of the spring and what is the effect of the error of that measurement on your determination of the spring constant?"
Do I just write +/- 0.5 once I average out the k values?

 

[haruspex]

First, k cannot be negative. Check your reasoning there.
Second, are you taking the average computed k as the answer?
The complete answer to your question is rather complicated. Because you divide by the displacement, a +/-0.5 cm error has a bigger impact on the answer at a small displacement than at a larger one. E.g. for your 5cm displacement that's a +/-10% error range.
A better procedure might be to sum the masses across the trials and divide by the sum of the displacements.
Your answer will also depend on the theory you have been taught. Does it include a root sum square formula?

 

[RJLiberator]

First, k cannot be negative. Check your reasoning there.

Good point. The negative from the direction of force must cancel the initial negative.

The complete answer to your question is rather complicated. Because you divide by the displacement, a +/-0.5 cm error has a bigger impact on the answer at a small displacement than at a larger one. E.g. for your 5cm displacement that's a +/-10% error range.

Yes, this is what I was strugglin with on how to proceed.

A better procedure might be to sum the masses across the trials and divide by the sum of the displacements.

This would be the average, correct? (.25+.1+0.05)/(.24+.1+.05) = 1.026.

I guess, I'm strugglin on how to apply the error measurement as instructed. We've only had one class and this is due the next time we meet and we haven't mentioned anything regarding errors, so it is just assumed. I have been taught the root sum square formula before.

The homework states:

Estimate the error on this measurement. For example, how accurately can you measure the displacement of the spring and what is the effect of the error of that measurement on your determination of the spring constant? Your final answer for the spring constant should be in the form k ± Δk, where Δk is the error.

So I am not entirely sure how to intrepret that.
I understand that the error on measurement is +/- 0.5 cm.

 

[haruspex]

This would be the average, correct? (.25+.1+0.05)/(.24+.1+.05) = 1.026.

Yes, but don't forget g.
Based on what you have been taught, what can you say about the error in the sum of the distances?

 

[RJLiberator]

Yes indeed. so 1.026*9.8 = 10.05 N/M as the average answer.

Hm. The error in the sum of the distances. I would think that we are working with +0.5 cm --> +0.005m would be 3*0.005 = +/-0.015m as the error. This is the propagation of errors.

So we know can say (.25+.1+0.05)/(.24+1+.05 with an error of +/- 0.015)

EDIT: Found the propagation of error with respects to division:

0.015/(0.24+.1+0.05) = 0.038
= 3.8%
So we take sqrt(3.8^2) = 3.8% is the uncertainty in answer.

Agree?

 

[RJLiberator]

10.05 N/M +/- 0.3819 would be the answer using this method. As 0.3819 is 3.8% of 10.05

 

[haruspex]

005m would be 3*0.005 = +/-0.015m as the error. This is the propagation of errors.

That is the extreme possibility, but very unlikely. It is more usual (though I have personal reservations about this) to apply the root sum squares formula to find the likely error range for the sum.

 

[RJLiberator]

Root sum squares formula.

So we take the square root of the sum of the errors and square it?
sqrt(0.015^2) = 0.015 would be the error then? 1.5% using the RSS method.

 

[haruspex]

Root sum squares formula.

So we take the square root of the sum of the errors and square it?
sqrt(0.015^2) = 0.015 would be the error then? 1.5% using the RSS method.

No, the method is square the errors, sum those squares, then take the square root. That's why it is called root-sum-square.
Since all three are .005m, we can take that out as a factor, making it a little simpler.

 

[RJLiberator]

Ah, I see. So with that method we get 0.009 * 100% = 0.9% error.
Then I would do 10.1*0.009 = 0.091 so the final answer would be 10.1+/-0.091

I thank you, greatly, for your help. Is there any final words you can say on the error calculation here?
We have used two different methods and arrived at different error results.
What method is better? Does it matter in regards to this problem?

 

[haruspex]

Ah, I see. So with that method we get 0.009 * 100% = 0.9% error.
Then I would do 10.1*0.009 = 0.091 so the final answer would be 10.1+/-0.091

I thank you, greatly, for your help. Is there any final words you can say on the error calculation here?
We have used two different methods and arrived at different error results.
What method is better? Does it matter in regards to this problem?

Which other method are you referring to? Do you mean the one at the end of post #1? Pardon my saying so, but I would not have honoured that with the description "method". Wild guess, perhaps.
I mentioned a personal reservation about applying RSS in this context. RSS presumes the source errors represent some fixed number of standard deviations in a roughly Gaussian distribution. It calculates the error range for the same number of standard deviations in the sum of the source values. The number of sd's doesn't matter as long as it is the same for all source values and understood to apply to the answer too.
But in this case,as often, the source errors have uniform distributions. Worse, the error in the sum will not have a uniform distribution. So in principle the quoted error range for the answer ought to be qualified by stating the number of sd's it represents.
Hope you followed that.

 

[RJLiberator]

Thank you for this analysis. I think you've armed me with enough information to tackle this problem.

Cheers.