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

In summary, the conversation discusses using experimental values for mass and period of oscillations to graph period^2 against mass in order to find the spring constant, with a desired gradient of 4(pi^2)/k. However, when using the equation and data, the calculated value for k is incorrect. After further investigation, it is discovered that the two data points used were too close together, resulting in a potentially inaccurate slope. It is suggested to use data points with a larger difference in period to get a more accurate slope and ultimately, a more accurate value for k.
  • #1
crankine
5
0

Homework Statement



Using experimental values for mass and period of oscillations, I'm trying to graph period[tex]^{2}[/tex] (y-axis) against mass (x-axis) 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.

Homework 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


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.
 
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  • #2
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
(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
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)
 
  • #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 that's 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
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
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 :)
 

1. What is Simple Harmonic Motion and how does it relate to vertical springs?

Simple Harmonic Motion is a type of periodic motion in which an object moves back and forth around a central equilibrium position. In the case of vertical springs, the object attached to the spring will oscillate up and down around its resting position, creating a sinusoidal motion.

2. What is the equation for calculating the period of a vertical spring's oscillation?

The equation for calculating the period of a vertical spring's oscillation is T = 2π√(m/k), where T is the period in seconds, m is the mass of the object attached to the spring in kilograms, and k is the spring constant in newtons per meter.

3. How does changing the mass attached to a vertical spring affect the period of its oscillation?

The period is directly proportional to the square root of the mass, meaning that as the mass increases, the period of oscillation will also increase. This is because the heavier the mass, the more force is needed to stretch or compress the spring, resulting in a longer period of oscillation.

4. Can we graph the period squared versus the mass of an object attached to a vertical spring?

Yes, we can graph the period squared versus the mass attached to a vertical spring. This graph will result in a straight line with a positive slope. The slope of the line will be equal to 4π²/k, where k is the spring constant. This relationship can be used to determine the spring constant of a vertical spring experimentally.

5. How can we use the graph of period squared versus mass to understand the behavior of a vertical spring?

The graph of period squared versus mass can help us understand the relationship between the mass of an object and the period of its oscillation on a vertical spring. It allows us to visualize how changes in mass affect the period and can also be used to determine the spring constant of the spring. Additionally, the graph can help us identify if there are any experimental errors in our data by comparing the slope of the line to the theoretical value of 4π²/k.

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