Question

I would like your feedback on my experiment. I am Investigating how different springs extend when I add different weights to it.

I want to check if the methodology of my experiment is good. Could you give me a feedback?

Methodology:

I have 3 springs of the same material but with different diameters/lenghts and 5 different weigths.

1. I will measure the initial length of the spring using a ruler, I will hang the weight under the spring. When the spring stop moving I will measure the new length. I will calculate the extension :ΔExtension = F inal length − Initial length

2. calculation

Averagelength=(L1 +L2 +L3)/3

Uncertainty = Range/2

I used a digital scale,The uncertainty of the mass is 0.005kg fort the uncertainty of the weight I multiplied by 10 the uncertainty of the mass.

During the experiment I noticed that the diameter will affect much more the spring constant than the size.

Answer 1

You want to measure the spring constant of several springs if i am correct. Firstly, what i would do is set up the anylitical equation for the spring constant, that is, apply Newton's second law on the mass when is hanging on the spring. Define downward as +y axis, then

$\:$

$\sum F_{y}=mg-k\Delta L=0$

$\Rightarrow \: \: \: mg=k\Delta L$

$\Rightarrow \: \: \: k=\frac{mg}{\Delta L}$

$\Rightarrow \: \: \: k=\frac{mg}{L_{f}-L_{i}} \: \: \: \: \: \: \: \: \: (1)$

$\:$

where,

$\:$

$m:\: \: mass$

$g:\: \: 9.8\, m/s^{2}$

$L_{f}:\: \: Final\: length\: of \: spring$

$L_{i}:\: \: Initial\: length\: of \: spring$

$k:\: \: spring\: constant$

$\:$

Second step is determine the uncertainty of the initial/final length. There are several statistical tools to determine the uncertainty of an average value, but ill go with one called "average deviation". Suppose you take 5 measurements for the initial/final length of the spring, then the uncertainty will be the average of all deviations. Consider the following example,

$\:$

# trial	Initial length, Li (cm)	deviation (cm)
1	5.1 cm	0.2 cm
2	5.4 cm	0.1 cm
3	5.6 cm	0.3 cm
4	5.2 cm	0.1 cm
5	5.3 cm	0 cm
average1	5.3 cm	0.1 cm

$\:$

Here is how i filled up this table,

$\:$

$average_{1}=\frac{5.1\, cm+5.4\, cm+5.6\, cm+5.2\, cm+5.3\, cm}{5}=5.32\, cm\approx 5.3\, cm$

$deviation_{1}=\left | 5.3\, cm-5.1\, cm \right |=0.2\, cm$

$deviation_{2}=\left | 5.3\, cm-5.4\, cm \right |=0.1\, cm$

$deviation_{3}=\left | 5.3\, cm-5.6\, cm \right |=0.3\, cm$

$deviation_{4}=\left | 5.3\, cm-5.2\, cm \right |=0.1\, cm$

$deviation_{5}=\left | 5.3\, cm-5.3\, cm \right |=0\, cm$

$AveDeviation=\frac{0.2\, cm+\, 0.1\, cm+0.3\, cm+0.1\, cm+0\, cm}{5}=0.14\, cm\approx 0.1\, cm$

$\:$

Thus, your initial length including uncertainty would be

$\:$

$L_{i}=5.3\, cm\pm 0.1\, cm$

$\:$

In a similar way, you get the final length of the spring including uncertainty. Suppose you get,

$\:$

$L_{f}=7.8\, cm\pm 0.1\, cm$

$\:$

Suppose the mass of your first weight is 100 g, then the next step is plug in all data into eq. (1) with their respective uncertainties,

$\:$

$k_{1}=\frac{(0.100\, kg\pm 0.005\, kg)(9.8\, m/s^{2})}{(0.078\, m\pm 0.001\, m)-(0.053\, m+0.001\, m)}$

$\:$

For substraction of two numbers with uncertainties, we use the formula

$\:$

$(x\pm \Delta x)-(y\pm \Delta y)=(x-y) \, \pm (\Delta x + \Delta y)$

$\:$

thus,

$\:$

$(0.078\, m\pm 0.001\, m)-(0.053\, m+0.001\, m)=(0.078-0.053)m\pm (0.001m+0.001m)$

$\:$

or

$\:$

$(0.078\, m\pm 0.001\, m)-(0.053\, m+0.001\, m)=0.025m\pm 0.002m$

$\:$

When multiplying a constant $c$ for a number with uncertainty, we use

$\:$

$c (x\pm \Delta x)=cx\pm c\Delta x$

$\:$

thus,

$\:$

$(0.100\, kg\pm 0.005\, kg)(9.8\, m/s^{2})=(0.980\pm 0.049)\, N$

$\:$

so far, k₁ is

$\:$

$k_{1}=\frac{(0.980 \pm 0.049)N}{(0.025\pm 0.002)m}$

$\:$

Formula for division between two numbers with uncertainty is given by

$\:$

$\frac{x\pm \Delta x}{y\pm \Delta y}=\frac{x}{y}\pm \left ( \frac{x}{y} \right )\left ( \frac{\Delta x}{x}+\frac{\Delta y}{y} \right )$

$\:$

thus,

$\:$

$k_{1}=\left [39.2\pm (39.2)(0.05+0.08) \right ]N/m$

$\:$

or

$\:$

${\color{Blue} k_{1}=(39\pm 5)\, N/m}$