B Why is my experimental spring constant an order of magnitude off?

AI Thread Summary
The discussion centers on a physics experiment comparing the maximum height of a projectile launched by a spring to the spring's compression. The experiment yielded a spring constant of 500 N/m, significantly lower than the expected 5000 N/m, leading to a 90% error. The user identified that the spring was not fully decompressed during the experiment, which likely contributed to the discrepancy. Additionally, factors such as the mass of the ball relative to the spring and various measurement errors were discussed as potential sources of error. The user expressed concern about the acceptability of the 90% error and received feedback on evaluating experimental accuracy.
Liddleton
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In my Physics IA, I investigated the relationship between the MAXIMUM height reached by a projectile (launched by a spring) and the COMPRESSION of that spring.

To evaluate my results I compared a spring constant, found using the relationship, against the spring constant found directly

However, my experimental spring constant differs (almost exactly) by an ORDER OF MAGNITUDE from the directly measured value. The cause of this percentage error is unclear, any help would be much appreciated!
I did my Physics IA on comparing the maximum height reached by a projectile (shot by a spring) and the compression of that spring.

In my experiment I found the maximum height reached by a metal ball shot directly upwards in comparison to the amount the spring was compressed.

To test the accuracy of my results, I compared the spring constant found through my experiment to the spring constant found directly (measuring the force needed to compress the spring over a certain distance).

To find the spring constant using my experiment data, I set the gravitational potential energy of the ball at its maximum height equal to the elastic potential energy:
π‘šπ‘”β„Ž=(1/2)π‘˜(π‘₯^2)
where:
π‘š = mass,
𝑔 = acceleration of gravity,
β„Ž = height of ball,
π‘˜ = spring constant,
π‘₯ = compression of spring.

Rearranging for π‘˜
2π‘šπ‘”β„Ž=π‘˜(π‘₯^2)
2π‘šπ‘”β„Ž/(π‘₯^2)=π‘˜
π‘˜=(2π‘šπ‘”)(β„Ž/π‘₯^2).

I separated β„Ž/(π‘₯^2) because it is the slope of the graph of the maximum height reached (y-axis) against the spring compression squared (x-axis), which I found.

Using these methods, experimentally, I got a value of π‘˜=500 N/m, whereas for my spring constant found directly (Force/compression of spring) I got a value of 5000N/m.

Can anyone help me try understand why there is such a big difference?

My found value is 90% off from the value which I'm expected to get and I can't think of anything that would introduce that much error.

If anyone wants the data:

The mass of the ball is 10.67g = 0.01067kg

and the value for the slope of β„Ž/π‘₯2 = 2500

IMPORTANT: If the mass of the ball was 100 grams instead of 10 I would get almost completely accurate results. But unfortunately I confirmed that the ball does in fact weight 10 grams, and so this is not the cause of uncertainty.

One last thing, If anyone happens to be an IB physics or science teacher in general, would you say 90% error is too large to take the data as acceptable? I know that the data doesn't have to be accurate but 90% feels like too much to me.

UPDATE: Im pretty sure I found the problem. With the apparatus I used to launch the ball, the spring is never fully decompressed, so the spring constant value I was finding was actually incorrect, I will try find a solution which allows me to find the correct spring constant directly - hopefully without having to dismantle the spring launcher.
 
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Liddleton said:
the value for the β„Ž/π‘₯2
= 2500.
What units?

Liddleton said:
β„Ž/(π‘₯^2) ... is the slope of the graph of the maximum height reached (y-axis) against the spring compression (x-axis),
This is incorrect. You can see that ##β„Ž/π‘₯^2## and the slope of this graph have different units.
 
Thanks for pointing out my minor mistakes, unfortunately they are in my conveying of the issue, and not in my processing of data. I meant to say the 'slope of the graph of the maximum height reached (y-axis) against the spring compression squared (x-axis) since this gives me a linear equation.
 
Liddleton said:
If the mass of the ball was 100 grams instead of 10 I would get almost completely accurate results.
A factor that could account for some of your observations is the relative mass of the ball and the spring. From the SMI Handbook of Spring Design: "These equations assume that the spring is massless and should only be used when the spring mass is less than 1/4 of the mass to be accelerated." That book then discusses calculations for accelerating loads that are large relative to the mass of the spring.

Liddleton said:
One last thing, If anyone happens to be an IB physics or science teacher in general, would you say 90% error is too large to take the data as acceptable? I know that the data doesn't have to be accurate but 90% feels like too much to me.
You need to compare the error to the estimated error budget. The error budget is the total of all errors. Some errors include:
Measurement error for peak height
Measurement error for spring rate
Friction between spring and housing
Measurement error for amount of spring compression
Measurement error for spring fully extended position
Correction factor for ratio of mass of ball to mass of spring
And more...

When your measured results are within your estimated error budget for at least two different weight balls, you probably have found all of the most significant error sources. Probably.
 
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Likes Dale and Liddleton
thanks you for the help! Luckily I think I've found the mistake, which was just due to a false assumption I made, luckily it is an easy fix! Nonetheless, thank you for the clarification on the evaluation for my IA, you helped me form a clearer understanding of what goal I am trying to reach in my evaluation.
 
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