How to find the value of a constant experimentally?

In summary, the conversation discusses finding the value of a constant experimentally using non-linear equations. The suggested method is to plot the experimental data and use the slope of the best-fit line to determine the value of the constant. Other methods, such as using Excel, are also mentioned but may not be as accurate. The importance of considering error and not blindly trusting computer-generated results is also emphasized.
  • #1
happyparticle
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TL;DR Summary
How to linearized an equation to find the slope
Hi,
First of all, sorry if this is not the right place to post my question I was not sure where exactly to post this kind of question.

I'm wondering how can I find the value of a constant experimentally.
For instance, I have a equation ##l = AB^{4/3}##, with a set of data for ##I## and ##B##.
If the equation above was linear I could find the slope of the graph to get the value for ##A##.

However, this is not a linear equation.
I wondering if I have to linearize the equation and then find the slope to get the value for ##A##?
If so, how exactly I linearize it?

For instance an equation like ##A = BC^{2}e^{-D/C}## could be linearized by multiplying both side by ln, I think.

Thank you
 
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  • #2
You could always plot I on one axis and [itex]B^{4/3}[/itex] on the other.
 
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  • #3
Vanadium 50 said:
You could always plot I on one axis and [itex]B^{4/3}[/itex] on the other.
Alright,
One more thing, in my case, what exactly linearized the equation means?
 
  • #4
I must reformulate my question.
I have a set of value for ##l## and ##B## which gives me a curve when I plot it and I'm confident that this is the right curve.
However, I would like to plot the theoretical equation which is ##l=AB^{3/2}## only for low value of ##B##.
I'm not sure how to do that. This is probably why my question was confusing.
I think I have to find ##A## first then plug ##A## with my values of ##B## to find ##l## ? I'm not sure if that makes sense.
 
  • #5
Here is what I would do.
Plot the experimental values of I vs B3/2. Put estimated error bars on each plotted point (I, B3/2). Draw the best straight line (yes use a ruler) through the points: the slope will equal A.
Use this value for A to create the curve over the region you desire.

If I knew more about the data I might use a different method., but you are on the right track. The error bars are a useful guide.
 
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  • #6
hutchphd said:
Here is what I would do.
Plot the experimental values of I vs B3/2. Put estimated error bars on each plotted point (I, B3/2). Draw the best straight line (yes use a ruler) through the points: the slope will equal A.
Use this value for A to create the curve over the region you desire.

If I knew more about the data I might use a different method., but you are on the right track. The error bars are a useful guide.
If you don't like the ruler method, use Xcel to create a table with I and B3/2, have it determine the linear regression coefficients in that table to get A. Then create a new table of I in one column and AB3/2 in the other and have it make a chart that way.
 
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  • #7
ohwilleke said:
If you don't like the ruler method
I prefer it when there are error "rectangles" on the data (EXCEL doesn't do that easilly as far as I know) You eyeballs do a remarkable good "RMS" fit. But I do (or at least did) love EXCEL for data. Saved me hundreds of man-hrs.
 
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  • #8
Yeah, I would rather hand-draw it than Excel. Hand-drawing gives you perspective, so it's a better place to start. Also, the error analysis doing this is not quite right, and Excel makes things look more certain than they actually are. After all, it came out of a computer - what could possibly be wromg?
 
  • #9
Vanadium 50 said:
After all, it came out of a computer - what could possibly be wromg?
"Garbage in, Gospel out"
 
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