# Spring Constant off by a factor of two

1. Nov 16, 2009

### rakbarut

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!!!)

2. Nov 16, 2009

### diazona

Re: Spring Constant off by a factor of two!!

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. Nov 16, 2009

### rl.bhat

Re: Spring Constant off by a factor of two!!

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.

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