Spring Constant off by a factor of two

In summary, the conversation was about a lab where the objective was to find the spring constant of a spring scale. The calculations done were off by a factor of two from the expected answer based on Hooke's Law. After further discussion, it was determined that the discrepancy was due to the mass of the object in motion and the loss of energy in the form of heat during the oscillations of the spring and mass. The correct expression for the work done on the spring is mg=1/2*k*x.
  • #1
rakbarut
10
0
Spring Constant off by a factor of two!

Hey everyone, I am doing a lab in which the objective is to find the spring constant of a spring scale, however when doing my calculations, the number I got was off by a factor of two from the supposed answer calculated from Hooke's Law. Here's how I did it...

So we have a spring scale, a 100g mass, and a meter stick. I attached the spring scale to the wall so it was secure and free to be used. I added the 100g mass and measured both the spring displacement and the force. I got x=.0047 m and F=0.9 N. So by using the law of conservation of energy, and setting the maximum stretch of the spring as my zero, I found that...

mgx=1/2kx^2 since all the grav. potential energy got converted to elastic potential energy.

However, manipulating the problem yields k=2mg/x, not k=mg/x as Hooke's Law proposes...

Please any help would be much appreciated! (I under a time crunch as well!)
 
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  • #2


Actually, if you were really watching carefully, you'd see that when the mass had dropped by a distance x, it would also be moving. Its inertia would carry it down by a further distance of x, for a total of 2x, before the spring force was able to stop it. Of course, when it got down to that 2x displacement, there would be an excess upward force on it from the spring, so it would go back up and stop again at its original position, then drop again, etc. etc. etc. (This is called simple harmonic motion, if you want to look up more information about it)

In reality, what happens is that the spring/mass loses some energy (in the form of heat, kind of like friction) every time it goes through this up-and-down oscillation, so it has less energy to move over time and the oscillations get smaller and smaller. Depending on what kind of spring you've got, the time it takes for the oscillations to become so small you can't tell they're there could be very quick, or it could take a long time.
 
  • #3


mg = 0.1*g =...?
In the case of Hook's law, the work done on the spring is completely stored in the spring as potential energy.
In the given problem, there is a rise in potential energy of teh spring and decrease in the of the block. So mg = 1/2*k*x is the correct expression.
 

Related to Spring Constant off by a factor of two

What is the "Spring Constant off by a factor of two"?

The "Spring Constant off by a factor of two" refers to a common error in physics experiments where the measured value of the spring constant is twice the expected value. This can occur due to various factors such as equipment limitations, human error, or incorrect calculations.

How does the error in spring constant affect the results of an experiment?

The error in spring constant can significantly impact the results of an experiment, as it affects the accuracy of calculations and measurements. It can lead to incorrect conclusions about the relationship between variables and affect the overall validity of the experiment.

What are some possible reasons for the "Spring Constant off by a factor of two" error?

There can be several reasons for this error, including using an incorrect formula to calculate the spring constant, using an inaccurate measuring device, or not accounting for external factors such as friction. Human error, such as incorrect data entry or reading measurements incorrectly, can also contribute to this error.

How can the "Spring Constant off by a factor of two" error be corrected?

The first step to correcting this error is to identify the source of the problem. This may involve double-checking calculations, calibrating equipment, or adjusting for external factors. It is also essential to carefully review and verify data to ensure accuracy. Conducting multiple trials and taking the average of the results can also help to reduce this error.

How can I prevent the "Spring Constant off by a factor of two" error in my experiments?

To prevent this error, it is crucial to carefully plan and conduct experiments, follow proper procedures, and use accurate and calibrated equipment. Double-checking calculations and having a colleague review the data can also help to catch any potential errors. It is also essential to be mindful of the expected value of the spring constant and make adjustments if the measured value deviates significantly from it.

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