Measuring g with a spring without knowing the mass

1. Oct 30, 2011

charbon

I have taken the challenge to measure the earth's gravitational force, g, without knowing the mass of an object. To do this, we took a spring and had it oscillate vertically and determined the period T.

Here is my theoretical development:
F = -kx
x(t) = Acos(ωt+θ) where ω = √(k/m)

of course, ω is the angular frequency and can be written as 2π/T

so let's isolate m.

m = k*T^2/4π^2

g = -kx/(k*T^2/4π^2)

and therefor g = 4π^2*x/T^2

That's a very nice formula until you end up testing it. With a little work, we determined the measurement of g was always off by a factor of 1/T and therefor the formula should be g = 4π^2*x/T^3.

Where does this extra T come from? We've been trying to figure this out for a long time.

2. Oct 30, 2011

Dickfore

This would imply that the gravitational force is in equilibrium with the elastic force. Therefore, the "x" in this formula is the elongation of the equilibrium position of the spring.

3. Oct 30, 2011

BruceW

If we had a spring with zero gravity, the total force on the mass would be:
$$F = -kx$$
(where $x_0=0$ is the equilibrium point of the spring)
But if there is gravity in the negative x direction, then we must put this force in as well:
$$F = -kx -mg$$
So the new equilibrium point is when
$$x_0 = - \frac{mg}{k}$$
(because when x equals this, the total force on the mass equals zero).
We can now make the equation simpler by introducing a new variable $q = x - x_0$, and we substitute this into our equation for the force:
$$\frac{d^2(q+x_0)}{dt^2} = -k(q+x_0) - mg$$
And, using our value for $x_0$, (and seeing that this is a constant), and doing some rearranging, we get:
$$\frac{d^2q}{dt^2} = -kq$$
So, in other words, we have F = -kq So it is the same physics, except that the equilibrium point has been shifted downwards by an amount equal to:
$$\frac{mg}{k}$$

In other words, studying the motion of the spring will not tell you the value of g. The only way to find g would be to measure how far the equilibrium point has been shifted by. (And this does require knowing the mass).

4. Oct 30, 2011

charbon

I should have added that the only reason I was studying the motion of the spring was to get the period T. I was indeed considering x to be the equilibrium position and I'm sure we can determine g without knowing the mass as experimentally, g = 4π^2x/T^3 as I said in the earlier post. I have tried with 5 different masses and this equation holds true for all of them.

5. Oct 30, 2011

Dickfore

This equation is not dimensionally consistent. Therefore, it cannot be correct.

Please post some raw data so that we can see what's going on.

6. Oct 30, 2011

charbon

m (kg) x (m) T (s) g (m/s^2) g/9.8
0.01 0.010 0.34 3.41 0.35
0.02 0.035 0.52 5.10 0.52
0.03 0.064 0.63 6.36 0.65
0.04 0.092 0.72 7.00 0.74
0.05 0.122 0.80 7.52 0.77

7. Oct 30, 2011

Dickfore

If you look at the last column, you can clearly see that your estimation of g/9.8 rises systematically with increasing mass. This should not be the case! I would say you probably made some wrong measurement of the displacement.

8. Oct 30, 2011

charbon

Notice the last column looks very similar to the third column. g has not been constant with our measurements and that is what the problem has been. We've measures and remeasured the displacement but nothing seems to be off. We also checked the spring constant to make sure it did not change by plotting F by x and it stayed linear. I really have no idea what is going wrong with our system.

9. Oct 30, 2011

Dickfore

There are two things you can check:
a) $x_{0} = \frac{m \, g}{k} \propto m$

b) $T = 2 \pi \sqrt{\frac{m}{k}} \Rightarrow T^{2} \propto m$

Using your data, I plotted and fitted the above (see the attached https://www.physicsforums.com/attachment.php?attachmentid=40516&stc=1&d=1320014248" document). It seems your dependence:
$$\frac{x}{1 \, \mathrm{m}} = 2.81 \, \frac{m}{1 \, \mathrm{kg}} - 0.0197, \ R^{2} = 0.999102$$
contains a significant constant term (twice the value of the first "equilibrium displacement" and more than half of the second).

I would say, you included both the amplitude and the equilibrium displacement. DId you stretch the spring arbitrarily before measuring the period? Then, did you look at the maximum position of the weight? If yes, your determination of x is definitely incorrect and you need to repeat the procedure.

Attached Files:

• gfromspring.xls
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10. Oct 30, 2011

charbon

I measured x by measuring first the spring without a mass on it using a ruler. I then put the mass on it and measured the length of the spring at its equilibrium. I then subtracted these two measurements for the result.

11. Oct 30, 2011

Dickfore

How did you make the system oscillate?

12. Oct 30, 2011

charbon

I pulled the mass down a bit and let it go.

13. Oct 30, 2011

Dickfore

Assuming you did everything correctly, the slopes of the two linear functions give:

$$\frac{g/k}{\mathrm{m}/\mathrm{kg}} = 2.81$$

$$\frac{4 \pi^{2}/k}{\mathrm{s}^{2}/\mathrm{kg}} = 12.968$$

This gives the following estimates for the spring constant (assuming $g = 9.8 \, \mathrm{m}/\mathrm{s}^2$):
$$k_{1} = \frac{9.8}{2.81} \, \frac{\mathrm{N}}{\mathrm{m}} = 3.5 \, \frac{\mathrm{N}}{\mathrm{m}}$$

$$k_{2} = \frac{4 \times 3.142^2}{12.968} \, \frac{\mathrm{N}}{\mathrm{m}} = 3.04 \, \frac{\mathrm{N}}{\mathrm{m}}$$

Even with your experimental uncertainty, I think this difference is significant, so you definitely have a systematic effect.

14. Oct 30, 2011

Dickfore

How did you measure the period?

15. Oct 30, 2011

charbon

we measured the spring constant to be around 3.33 (I don't remember the standard deviation on that number but I think it was around 0.1)
I'll probably return to the lab in the next couple of days to get more measurements.

16. Oct 30, 2011

charbon

The period was determined using a position sensor. We plotted x(t) and established the time it took for the system to oscillate 20 times. We then divided that number by 20.

17. Oct 30, 2011

Dickfore

I also see that the region of your displacements spans one order of magnitude (from 1 cm to 12 cm). Is it possible that you went outside of the region where Hooke's Law applies?

EDIT:

Assuming $F_{\mathrm{el}} = k_{n} \, x^{1 + n}$, then the equilibrium elongation would be:
$$x_{0} = \left( \frac{m \, g}{k_{n}} \right)^{\frac{1}{1 + n}}$$
Taking the log of both sides:
$$\log{\left( \frac{x_{0}}{1 \, \mathrm{m}} \right)} = A + B \, \log{ \left( \frac{m}{1 \, \mathrm{kg}} \right)}, \ A = \frac{\log{ \left( \frac{g/k_{n}}{\mathrm{m}^{n}/\mathrm{kg}} \right) }}{n + 1}, \ B = \frac{1}{1 + n}$$

Using the same data and updating the Excel file, we get:
$$\log{\left( \frac{x_{0}}{1 \, \mathrm{m}} \right)} = 1.144 + 1.556 \, \log{\left( \frac{m}{1 \, \mathrm{kg}} \right)} \Rightarrow 1.556 = \frac{1}{1 + n} \Rightarrow n = \frac{1}{1.556} - 1 = -0.357$$

It seems your spring does not obey Hooke's Law!

Attached Files:

• gfromspring.xls
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Views:
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Last edited: Oct 30, 2011
18. Oct 30, 2011

BruceW

I just looked back on this thread and realised I was wrong. charbon, you were right that theoretically,
$$g=\frac{4 \pi^2 x}{T^2}$$
Where x is the shift of the equilibrium position due to the effect of gravity.

Its a shame your results don't agree with theory. The problem with springs is that they only obey Hooke's law within a small range of masses and displacements. Also I am not sure that measuring the extension with no attached masses will give a good indication of the equilibrium point without gravity. (Even though it should, in theory).

I suggest looking at the data and trying to guess what specifically might have caused the difference between observation and theory. I would try to have a look myself, but I'm pretty tired right now. I hope you manage to sort it all out!

19. Oct 30, 2011

Dickfore

If the spring does not obey Hooke's Law, then the oscillation period depends on the amplitude of oscillation!

20. Oct 30, 2011

charbon

I did measure the spring's constant for my largest mass and it's value seemed pretty constant. I'd like to take a closer look at this though so I will remeasure it. for the other masses though, it should not be a problem.