# Simple Harmonic Motion: vertical springs - graphing period^2 & mass

by crankine
Tags: harmonic, mass, motion, period, simple
 P: 5 1. The problem statement, all variables and given/known data Using experimental values for mass and period of oscillations, I'm trying to graph period$$^{2}$$ (y-axis) against mass (x-axis) to get a gradient = 4(pi$$^{2}$$)/k so I can find k (spring constant). The problem is that when I use each pair of results in the equation below, I get the right answer for k, but when I graph it and try to use the gradient its wrong!!!! I'm pretty confident I've substitued correctly etc. If you want the exact data then I would put it up but I don't know how to put tables. 2. Relevant equations T = 2pi√(mass/k): T$$^{2}$$ = 4(pi$$^{2}$$)(mass/k) -> compared to y=mx, m=4(pi$$^{2}$$)/k k = 4mass(pi$$^{2}$$)/(T$$^{2}$$) -> substituting each pair of T and mass here I get the right answer 3. The attempt at a solution I've tried swapping the axes and graphing T against m$$^{1/2}$$ and so on but nothing has got me any closer to the right answer.
 Mentor P: 12,074 Welcome to Physics Forums. That's weird, your equations and method all look correct. Here's what I suggest: pick any 2 data points, and post them here. Also post the slope you calculate based on just those 2 points. Then we can both look at your work and try to figure out what's going on.
 P: 5 (the right value for k is about 4 or 5) mass 0.06 period 0.7 period^2 0.49 calculated value for k 4.834091952 mass 0.1 period 0.8 period^2 0.64 calculated value for k 6.168502751 graph of period^2 (y) and mass (x) gives gradient 1.2558, value for k from this is 31.4... halp! :) thankyou
 Mentor P: 12,074 Simple Harmonic Motion: vertical springs - graphing period^2 & mass Thanks for posting the data. Here are questions and comments: What are the units on the masses? It's difficult to get an accurate calculation of the slope from the data you give here, because the two periods are so close together (0.7 s and 0.8 s). Do you have 2 data points where the periods are not so close to each other? If we do use those 2 data points to calculate a slope, we get (0.64 - 0.49) / (0.1 - 0.06) = ____ ?(Hint: it's not 1.2558)
 P: 5 Thanks :) All SI units so kg, seconds etc. I didn't calculate the slope myself, I put it into excel - I think it made the points (0.64,0.49) and (0.1,0.06) i've sorted that so thats okay now. BUT...... now the gradient = 3.75 now... so k = 10.5. This is still double what it should be, and double what I calculated!! argh.
 Mentor P: 12,074 This is looking better, at least the math is correct for those two points. You could easily be a factor of two high or low when the two periods, 0.7s and 0.8s, are so close to one another. How about 2 data points that are not so close together in period? That would be a lot closer to the actual slope.
 P: 5 No I don't I'm afraid as I only used 3 masses, so the period was 0.7, 0.75 and 0.8. I think you're right though because when I added a theoretical higher point that fitted with the equation the gradient improved, so I'll go back and repeat the experiment with smaller and greater masses. Thanks for your help :)

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