Lab Report
Jaehun Shim
Research question
Verifying the effect of the weight of the string on the time period of the bifilar pendulum and finding
the spring constant K.
Independent variable
Time period which is measured in 10 seconds.
Dependent variable
The mass of the spring that is changed, which is measured in kg.
Controlled variable
•Type of spring used
•Environment where the investigation was conducted
•The type of pendulum used.
Materials/apparatus used
•iPad Stopwatch
•Wooden ruler
•100 gram pendulum
•Bifilar Pendulum
Experimental Set-Up:
Procedure:
1. Place a bifilar pendulum on the flat table and hang the spring.
2. Hang 100 g pendulum on the string.
3. Pull the spring and release it.
4. Record time until the spring oscillate 10 times.
5. Add 100 g pendulum and repeat 3, 4.
Data Collected
Table 1; Time taken for 10 oscillations (s) for varying length of string (m)
Mass of
the
spring
(kg) ±
0.01 kg
Time taken for 10 oscillations (s) ± 0.01 s
Trial
Average
Absolute
uncertain
ty
1
2
3
4
5
6
7
0.1
2.1
2.2
1.8
2.0
1.8
2.1
1.9
2.0
0.3
0.2
3.1
3.0
3.2
3.3
3.4
3.5
3.2
3.2
0.3
0.3
4.0
4.1
4.1
3.9
3.9
4.0
4.0
4.0
0.1
0.4
4.8
4.8
5.1
4.9
4.8
5.0
5.0
4.9
0.3
0.5
5.4
5.3
5.6
5.6
5.5
5.5
5.6
5.5
0.2
The data above gives me the time period for 10 oscillation for the mass of spring from 0.1 kg to 0.5 kg.
This is to allow for more accuracy. The trials were repeated 7 times minimize human error in the
experiment. Absolute uncertainty was calculated in the repeated measurements by finding the 1/2(max
value – min value) in the spread of data.
Table 2; Mass of the spring (m) and the Time period Squared (T²) and the absolute uncertainty
Mass of the spring (kg)
± 0.01 kg
Average time taken for 1
oscillation (s) ± 0.01 s
Time period square T² (s² ) Absolute Uncertainty of T²
(s²)
0.1
0.2
0.04
0.3
0.2
0.3
0.1
0.2
0.3
0.4
0.2
0.05
0.4
0.5
0.2
0.1
0.5
0.6
0.3
0.07
The data above gives the data of the time period squared for 1 oscillation for the mass of spring from
0.1 kg to 0.5 kg which is found by time for 10 oscillations divided by 10 and the absolute uncertainty of
T². In order to ensure that the relationship between the time period, and the mass of the spring is linear,
it is important to graph T², in relation to mass (kg), as they are proportional to each other. If not, the
graph I would get will not be linear, but rather curved – showing a root relation.
Relationship between T² and mass
0.4
0.3
y = 0.66x - 0.03
R² = 0.9973
y = 1.3x - 0.28
R² = 1
y = 0.1x + 0.18
R² = 1
0.2
T^2 (s^2)
0.1
0
-0.1
-0.2
-0.3
-0.4
0
0.1
0.2
0.3
0.4
0.5
0.6
Mass (kg)
Presentation of the data
Blue line = Max line (y = 0.66x - 0.03)
Green line = Min line (y = 0.1x + 0.18)
Red line = Best-fit line (y = 1.3x -0.28)
From the graph, it seems that as the mass of the string increases, T² also increases and It has the
positive relationship between the mass of the string, and the time period. Given that the relationship is
linear in nature, it is appropriate to utilize or construct a maximum and minimum line also, which is seen
in the graph itself.
The original formula is 4π²/T² = k/m and I simplified it into T² = 4π²m/k.
Following to the formula T² = 4π²m/k, T² is Y and m is X. Therefore gradient is equals to 4π²/k. The
equation of the best-fit line is y = 1.3x - 0.28 so gradient of best-fit line is 1.3. I solved the equation 4π²/
k = 1.3 and I got 30.7 for K value which is spring constant.
K = 30.7