
#1
Apr1509, 10:56 AM

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[tex]^{2}[/tex] (yaxis) against mass (xaxis) to get a gradient = 4(pi[tex]^{2}[/tex])/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[tex]^{2}[/tex] = 4(pi[tex]^{2}[/tex])(mass/k) > compared to y=mx, m=4(pi[tex]^{2}[/tex])/k k = 4mass(pi[tex]^{2}[/tex])/(T[tex]^{2}[/tex]) > 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[tex]^{1/2}[/tex] and so on but nothing has got me any closer to the right answer. 



#2
Apr1509, 06:58 PM

Mentor
P: 11,988

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. 



#3
Apr1609, 06:00 AM

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 



#4
Apr1609, 01:04 PM

Mentor
P: 11,988

Simple Harmonic Motion: vertical springs  graphing period^2 & mass
Thanks for posting the data. Here are questions and comments:




#5
Apr1609, 05:43 PM

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. 



#6
Apr1609, 06:39 PM

Mentor
P: 11,988

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. 



#7
Apr1709, 06:13 AM

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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