Finding the Spring Constant

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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

*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 = sum of values = 3.81 + 3.63 + 3.74 = 3.73s
                     number                               3

To find the uncertainty for time, the biggest difference (regardless of whether it’s the maximum or minimum value) needs to be taken away from the average.  

3.81 – 3.73 = 0.08

3.73- 3.63 = 0.10s

So we use ± 0.10s for the first mass.

For the time of the trials, the uncertainty is 0.01(this being the equipment error) + reaction time.

So 0.01 + 0.2

= ± 0.21s

Processed Data:

Table 2- Processed data (Period, Uncertainties, T² and more stated in the table)

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:

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(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 ...

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