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Help with spring constant units

  • Thread starter NewSoul
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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?
 

Answers and Replies

  • #2
Simon Bridge
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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
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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
Simon Bridge
Science Advisor
Homework Helper
17,847
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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?
 

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