Simple Spring Constant Lab not adding up

In summary, the person is having issues with their spring constants, which seem to be too small. They provide data on masses and displacement, and mention an equation they used to calculate the spring constant. However, they realize their mistake in converting units and thank others for pointing it out.
  • #1
Animal
16
0
For some reason, my spring constants are all screwy. an example

Masses (Newtons) .049 .098 .196 .294 .490
Displacement (kg) .006 .019 .035 .055 .090

and the line i got was Mass=.188d-.001

doesnt that mean that the spring constant is .188 N/m? that doesn't make any sense though, because i did 4 other springs and the stiffest one had a spring constant of .067. what am i doing wrong?
 
Physics news on Phys.org
  • #2
Displacement is not Kg.

your intercept is wrong.The intercept should be the natural length of the spring, so it can't be negative.

You should redo the experiment.
 
  • #3
right, i meant meters...not kilograms. typo.


as far as the y intercept being off, i set it equal to zero, so the original length would be zero. besides, those are millimeters, and there's something big wrong with the calculations...
 
  • #4
Doubling your mass (such as from .049 to .098), should NOT more than double your displacement (which, respectively was .006 and then .019).

Also, those spring constants seem extraordinarily small. You are saying that using your stiffest spring, it only takes an increase of .067 N to increase the displacement by a meter. Do you remember your equations correctly? (hell, the units of a spring constant give it away)
 
  • #5
i know, that's the problem. i have no idea why my constants are so small. here, the original data collected was:

5g 10g 20g 30g 50g
0.6cm 1.9 3.5 5.5 9.0

i divided all the grams by 1000 to get the weight in kilos, then multiplied them by 9.8 to get the mass in Newton.
i divided all lengths by 100 to get the number in meters...

im sure my problem was in my conversion, because it simply can't be anything else. any ideas?
 
  • #6
Animal said:
i know, that's the problem. i have no idea why my constants are so small.

Just looking at your line, I'd say you've fitted the inverse line:
you seem to have written d = 0.188 M - 0.01 instead of M = 0.188 d - 0.01.

Fill in your data in your formula and you'll see: d is roughly about 1/5th of M, numerically.
 
  • #7
Umm, dividing (.005kg * 9.8m/s^2) / (.006m) results in 8.167. So are you dividing wrong or something? (obviously this isn't the average, but I am just showing that it works out for the first data point).
 
  • #8
vanesch, you were right. my axises (axes?) were reversed. thanks for the help guys
 

Related to Simple Spring Constant Lab not adding up

1. What is a simple spring constant lab?

A simple spring constant lab is a laboratory experiment that involves measuring the force required to stretch or compress a spring and using this data to calculate the spring constant, which represents the stiffness of the spring.

2. Why is my calculated spring constant not matching the expected value?

There could be several reasons for this discrepancy, including experimental error, inaccurate measurements, or equipment malfunction. It is important to carefully follow the instructions and perform multiple trials to reduce errors and increase the accuracy of your results.

3. Can I use any type of spring for this lab?

No, it is important to use a simple, linear spring for this experiment. Non-linear or complex springs may not follow Hooke's law, which is necessary for calculating the spring constant.

4. How many trials should I perform for this lab?

It is recommended to perform at least three trials for each measurement to reduce errors and increase the accuracy of your results. You can also perform more trials to ensure the consistency of your data.

5. What is the significance of the spring constant in real-life applications?

The spring constant is an important factor in various real-life applications, such as in the design of suspension systems, shock absorbers, and springs for different types of machinery. It helps engineers determine the appropriate stiffness of a spring for a specific application.

Similar threads

Replies
3
Views
1K
  • Introductory Physics Homework Help
Replies
4
Views
8K
  • Introductory Physics Homework Help
Replies
10
Views
3K
  • Introductory Physics Homework Help
Replies
4
Views
2K
Replies
1
Views
2K
  • Introductory Physics Homework Help
Replies
2
Views
2K
  • Introductory Physics Homework Help
Replies
3
Views
8K
  • Introductory Physics Homework Help
Replies
4
Views
2K
  • Introductory Physics Homework Help
Replies
8
Views
2K
Back
Top