how to find spring constant from graph
Find the spring constant 'k' by using the slope of my graph
- Thread starter physicsnewby
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Homework Statement
This should be straightforward but I'm not sure what to do. If I brand a graph and the Y axis is in cm, and the X is in g, does cm/chiliad make sense?
I'm trying to find the spring abiding 'm' past using the gradient of my graph. When I do this I become a huge number so I'thousand playing with units to bring it down.
Homework Equations
k = g/slope (m = 980 cm/s^2)
The Try at a Solution
using the info. gathered my k = (980 cm/s^ii) / 0.073 = 13520.55
this sounds extremely high. My slope units are what's throwing me I think. Does 0.073 cm/g make sense? And then, would the units be 13520.55 g/s^2 ?
I believe k is supposed to be in North k, so I'm not sure this is right ?!
Answers and Replies
You should give us the entire problem, exactly as it was given to you, and we can assist you with information technology.
- Warren
simply cheque it:
Potential energy stored in a compressed jump = (1/2)*k*x^ii ; where k is the spring constant and x is the compression in meters.
[J]=one thousand[thou^ii]
grand=[J/m^ii]=[Nm/yard^2]=[N/m]
The first graph is measuring displacement vs mass. So, in my case its cm vs grams.
The second is measuring menses squared (T^ii) vs mass. In my case, its seconds^squared vs grams.
I describe line of all-time fit and determine the slope. From here, K is determined using one of two equations. The kickoff graph is k=yard/slope, the second graph 4pi^2/slope.
Cartoon Max/Min lines for determining slope error, I go the mistake for gradient. Maybe this is where I become wrong.
I demand to do the math for discrepancy which is the difference betwixt the k values, divided past the square root of the sum of the squares of their errors (I used slope fault here - giving me a value around +/- 0.003)
My question most units comes here. Because my one thousand is huge (14000) and (15000) for example, when I do the discrepancy equation I get a ridiculous answer like 333333.33 when the answer should be close to 2.
I thought units may exist the problem for such a large value, just I'm not sure. The K'south I'm comparison will both have different values (graph ane: cm/g, graph 2: s^2/g) so how do I make them the same to compare?
Source: https://www.physicsforums.com/threads/find-the-spring-constant-k-by-using-the-slope-of-my-graph.174526/
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