- Level: International Baccalaureate
- Subject: Physics
- Word count: 2976
Finding the Spring Constant
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Introduction
Practical 2- The application of Hook’s law and SHM to calculate K (spring constant)
One example of simple harmonic motion is the oscillation of a mass on a spring. The period of oscillation depends on the spring constant of the spring and the mass that is oscillating. The equation for the period, T, where m is the suspended mass, and k is the spring constant is given as
We will use this relationship to find the spring constant of the spring and compare it to the spring constant found using Hooke’s Law.
- Independent variable: Mass (kg)
- Dependant variable: Time
DATA COLLECTION AND PROCESSING
Raw Data
Table 1 – Time needed for a spring to complete 10 oscillations
Mass, m/grams | Time for 10 oscillations (±0.21s) | ||||
Trial 1 | Trial 2 | Trial 3 | |||
100g ± 4 | 3.81 | 3.63 | 3.74 | ||
200g ± 8 | 5.46 | 5.30 | 5.37 | ||
300g ± 12 | 6.75 | 6.66 | 6.71 | ||
400g ± 16 | 7.78 | 7.81 | 7.78 | ||
500g ± 20 | 8.78 | 8.72 | 8.75 | ||
600g ± 24 | 9.69 | 9.60 | 9.62 | ||
700g ± 28 | 10.25 | 10.20 | 10.22 | ||
////////////////////////////// | /////////////////////////////// | //////////////////////////////// | /////////////////////////////// | ||
Qualitative observations | As more weights were added to the spring, the spring oscillated more slowly. However, when the mass surpassed 700g, the spring began to get deformed and its length changed therefore it could not be worked with. |
*The uncertainties for each mass will differ throughout because each weight is approximately 4g off, thus 2 weights will be 8g off, 3 weights will be 12g and so on. Generally, the slotted weights are not accurate, thus these uncertainties have been added.
Also, for Raw data we can add a similar note for ±0.21s. ie. ∆T= equipment error + reaction error
Example 1:
To obtain the uncertainties for time, the average needs to be found out using the following formula:
Average =
Middle
0.778
0.022
0.605
0.008
7.78± 0.03
0.500 kg
0.020
8.75
0.030
0.875
0.024
0.766
0.007
8.75± 0.03
0.600 kg
0.024
9.64
0.045
0.964
0.026
0.929
0.010
9.64± 0.05
0.700 kg
0.028
10.22
0.025
1.022
0.024
1.044
0.006
10.22± 0.03
The way we calculate the period T is basically by dividing the average time for each mass. The way we get the uncertainty of the period T is dividing the uncertainty gotten in the average period T and dividing that by 10 as well.
Example 2:
Primarily, what we do here is that we calculate the average error and they way we do that is simple. For instance, for the first mass (0.100 kg), this is how we calculate the averaging error:
(3.81 – 3.63) ÷ 2
=0.090
After that, we add this averaging uncertainty to 0.21 (which is basically the summation of our equipment and reaction time error)
Example 3:
Mass of 100g is 0.1kg as we divide by 1000. However to calculate the absolute uncertainty of mass in kg, we simply see how high the uncertainty was for the masses. For example, for the 100g mass, we weighed it and found that it was 90 something. After weighing all 7 masses, 96 was the lowest mass so 4g was the uncertainty. Thus it is ± 4g. For the next masses, it would be ± 8 and ± 12 and so forth.
Example 4:
To calculate the period, we divide average T by 10 since there were 10 oscillations. So lets say for the first mass 0.1 kg, the average T is 3.73. We divide 3.73 by 10 giving us 0.373.
What we do next is get ∆Period. So Basically what we do here is that we add all the timing uncertainties. ∆Equipment + ∆Reaction + ∆Average uncertainty
So for instance, again we use the first mass (0.100 kg) and the time taken for that as our example.
= 0.090 + 0.21
= > 0.300 After we have done this (∆Equipment + ∆Reaction + ∆Average uncertainty) we divide it by 10 in order to obtain ∆Period.
= 0.300 / 10 = 0.030 Therefore, ∆ Period for the first mass (0.100 kg) is 0.030
Example 5:
For calculating T², we square the average T. For example, for 0.1 kg, the average T is 0.373 s. So we multiply it by itself
So 0.373 x 0.373= 0.139 s
The formula for finding the spring constant is:
However, to obtain a straight line for the graph, both sides need to be squared, thus resulting in this formula being used:
*Note, to obtain a straight line we will be putting the formula in a “y=mx+c” form.
To find the uncertainties for T2, the percentage uncertainties need to be multiplied by 2.
To find the percentage uncertainty for time, the absolute uncertainty needs to be divided by the actual value. Since T2 has an exponent, the time needs to be doubled (multiplied by 2).
Uncertainty of T = 0.010 = 0.027
0.373
0.027 x 2 = 0.054
Note* the reason I am taking absolute values is because while I am at the graphing phase, I cannot use percentage values to graph uncertainties thus I will need the absolute values to graph uncertainties.
Table 3- Average times for each mass
Mass, m/kg | Time, T2/s2 |
0.100 ± 0.004 | 0.139 ± 0.054 |
0.200± 0.008 | 0.328 ± 0.034 |
0.300 ± 0.012 | 0.450 ± 0.015 |
0.400 ± 0.016 | 0.605 ± 0.008 |
0.500 ± 0.020 | 0.766 ± 0.007 |
0.600 ± 0.024 | 0.929 ± 0.010 |
0.700 ± 0.028 | 1.044 ± 0.006 |
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Conclusion
Weakness: Another weakness would be that damping might have taken place during our experiment which might have affected the accuracy of our results. The type of damping which might have occurred in our experiment would be light damping because we were working with air. In light damping, if the opposing forces are small, the result is a gradual loss of total energy. This means that the amplitude of the motion gets slowly less with time. For example, our mass on a spring hanging in the air would have a little damping due to air resistance.
Improvements: The way to avoid this weakness might be to take even more trials and work with more oscillations as that will reduce chances of error and we would get more accurate values and uncertainties in the end. Light damping due to air resistance normally occurs. However to avoid it, we should try and be more accurate to keep errors due to damping at its minimal. This we can do by working with more trials and oscillations. Also, we should include balancing with number of oscillations. Meaning that too many could result in more damping effects.
Lastly a very minor weakness could be the atmosphere we were working with. This is because the AC could have supplied the spring with more movement and motion. This could have caused a decrease in time and an increase in the spring constant k. The way to avoid this weakness is to basically turn off the AC and work with a normal atmosphere.
This student written piece of work is one of many that can be found in our International Baccalaureate Physics section.
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