Help with spring constant units

In summary, a student is trying to find the spring constant ks and the damping constant kd for a spring experiment. They first measured the displacement of the spring in a vertical position and found that ks = 25.3 ± 0.10 g/cm. They then lifted the spring to make it bounce and counted the number of oscillations to find kd = 24.1 kg/s2. However, they are unsure how to calculate the error for kd and are questioning the inclusion of the constant g in the equation.
  • #1
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Homework Statement


Part 1 (static method): We measured the displacement of a spring (in cm) after adding and subtracting masses (in g). The spring was placed in a vertical position. I am supposed to find the spring constant ks.

Part 2 (dynamic method): We did something similar to the above, but we lifted the spring a bit so it would bounce. We then counted the number of oscillations of the spring for 20 seconds each time we changed the masses put on the spring. I am then supposed to find kd.

Homework Equations


Hooke's law: F = kx
units of k = N/m (kg/s2)
ks = Sg

The Attempt at a Solution


I created a graph and found S + ΔS = 25.3 ± 0.10 g/cm

However, it's the equation ks = Sg that's messing me up. My instructions don't tell me what I'm supposed to do with g. I have no idea why it's stuck in there. According to this equation, ks ± Δks = 25.3 ± 0.10 g2/cm, but that doesn't seem to make any sense. How am I supposed to end up with the units of kg/s2 when I won't have any of those units until part 2? And if these are not the units I'm supposed to use, how do I know if I do have the correct units?
 
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  • #2
My instructions don't tell me what I'm supposed to do with g.
Don't you have an equation that tells you what to do with it?

How am I supposed to end up with the units of kg/s2 when I won't have any of those units until part 2
But you have all those units right there ... what are the units of g?

S is the slope of the graph for part 1?
It looks like you plotted mass on the vertical (y-axis) and extension on the horizontal (x-axis).
Is that correct?

Why not convert S to SI units - then use the standard value for g in SI units?
 
  • #3
Simon Bridge said:
Don't you have an equation that tells you what to do with it?

But you have all those units right there ... what are the units of g?

Ahh...it didn't mention in the instructions, but I just realized that g must mean the acceleration due to gravity. Gah! I knew it was something simple! With multiplying all of this by gravity, I end up with a much nicer 24.9 ± 0.10 kg/s2. Thanks!

S is the slope of the graph for part 1?
It looks like you plotted mass on the vertical (y-axis) and extension on the horizontal (x-axis).
Is that correct?
Yes, you are correct.

------

Perhaps you can help me in calculating an error for part 2? Or do I have to make a new thread?

Essentially...

My equation is kd = (4π2)/S. (This is a new slope with s2 on the y-axis and g on the x-axis.)
From the first equation, I found kd = 24.1 kg/s2.

I also have an equation for a function f(x) = Cxn, then... Δf/f = Cn(Δx/x). Δf is the error, C is a constant, and Δx/x is the fractional error in x.

Therefore, Δf must equal fCn(Δx/x). Now, it seems to me that this means Δkd = kd(4π2)(-1)(ΔS/S). However, this would cause Δkd to be -0.00190973753, which doesn't make sense to me. Shouldn't it be a positive number?

So it seems then that kd ± Δkd = 24.1 ± 0.0 kg/s2.
 
  • #4
NewSoul said:
Ahh...it didn't mention in the instructions, but I just realized that g must mean the acceleration due to gravity. Gah! I knew it was something simple! With multiplying all of this by gravity, I end up with a much nicer 24.9 ± 0.10 kg/s2. Thanks!

It sounds like you were following instructions without understanding the physics behind them.

When you hang a mass off the end of the spring, the the spring pulls on the mass with F=kg, and the mass pulls down with F=mg, so that mg=kx.

If you plot m vs x, you get a line with equation: m=(k/g)x
You could have plotted mg vs x, which would make the slope k.
Perhaps you can help me in calculating an error for part 2? Or do I have to make a new thread?

Essentially...

My equation is kd = (4π2)/S. (This is a new slope with s2 on the y-axis and g on the x-axis.)
From the first equation, I found kd = 24.1 kg/s2.

I also have an equation for a function f(x) = Cxn, then... Δf/f = Cn(Δx/x). Δf is the error, C is a constant, and Δx/x is the fractional error in x.

Therefore, Δf must equal fCn(Δx/x). Now, it seems to me that this means Δkd = kd(4π2)(-1)(ΔS/S). However, this would cause Δkd to be -0.00190973753, which doesn't make sense to me. Shouldn't it be a positive number?

So it seems then that kd ± Δkd = 24.1 ± 0.0 kg/s2.

if you have a measurement ##x\pm\Delta x##, and ##z=1/x##, what is ##\Delta z##?
Clearly it is not zero! That your problem?

It sounds like you are using equations without understanding the physics behind them.

well: xz=1 so $$\frac{\Delta x}{x}+\frac{\Delta z}{z}=\frac{\Delta 1}{1}\\
\Rightarrow \frac{\Delta x}{x}=\frac{\Delta1}{1}-\frac{\Delta z}{z}\\
\Rightarrow\frac{\Delta x}{x}=-\frac{\Delta z}{z}$$ ... since there is no uncertainty on a constant: Δ1=0.
What this does is tell you that you are right - it's not just you using the wrong equation.
It is important that you understand the equations that you use.

So - using your understanding of errors, how would you handle it?
What is the minus sign telling you in this case?

Hint: The error is always written out as ##\pm## - why?
 
  • #5


Hello, based on the information provided, it seems that you are on the right track in finding the spring constant ks in Part 1. The equation ks = Sg is correct, but the units may seem confusing at first. In order to convert from g/cm to N/m, you will need to use the conversion factor of 1 g = 0.0098 N and 1 cm = 0.01 m. This will give you the units of N/m for ks.

For Part 2, you will be using a different method to find the spring constant, which is why the units are different. In this case, the units for the spring constant kd will be in units of N/m/s^2. This is because in the dynamic method, you are measuring the frequency of the oscillations, which is in units of 1/s. Therefore, the units of kd will be N/m/s^2 or kg/s^2.

In summary, the units for the spring constant ks in Part 1 will be N/m, and for kd in Part 2, it will be N/m/s^2 or kg/s^2. It is important to pay attention to the units in order to ensure that your final answer is correct. I hope this helps. Good luck with your homework!
 

1. What is the standard unit for spring constant?

The standard unit for spring constant is Newtons per meter (N/m).

2. How do you calculate the spring constant?

The spring constant can be calculated by dividing the force applied to the spring by the resulting displacement.

3. Can spring constant be negative?

Yes, spring constant can be negative if the spring is compressed or stretched in the opposite direction of the applied force.

4. What is the difference between spring constant and spring stiffness?

Spring constant and spring stiffness are often used interchangeably, but technically spring stiffness refers to the force needed to stretch or compress a spring by a certain distance, while spring constant refers to the ratio of force to displacement.

5. How does temperature affect the spring constant?

Temperature can affect the spring constant by altering the elasticity of the material the spring is made of. In general, as temperature increases, the spring constant decreases due to increased molecular motion and less rigidity in the material.

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