- Pages: 3
- Word count: 747
- Category: Experiment
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Intent: To conduct an experiment to prove the yielding support distance is directly proportional to the period.
d(m)| Time for 20 Oscillations (s)| Time for 1 Oscillation T (s)|
T2(s2)| d3(m2) x 10-3| | 1| 2| 3| Average| | | |
0.24| 31.50| 31.47| 31.44| 31.47| 1.57| 2.46| 13.8| 0.21| 31.0| 30.97| 31.09| 31.02| 1.55| 2.41| 9.2|
0.l8| 30.56| 30.69| 30.69| 30.65| 1.53| 2.35| 5.8|
0.15| 30.44| 30.37| 30.20| 30.34| 1.52| 2.30| 3.4|
0.12| 30.16| 30.19| 30.22| 30.19| 1.51| 2.28| 1.7|
0.09| 30.00| 30.00| 29.96| 29.99| 1.50| 2.25| 0.7|
Interpretation: The lab was a success because it was proven that as length was of the length of the yielding support decreased, the period also decreased. A cause of error was due to human reaction time. Although 20 oscillations were counted, the person holding the stopwatch might have stopped milliseconds before or after, therefore causing the values to be slightly off. For this reason, an average was taken to increase accuracy. This could be prevented if the person who is counting the oscillations also times them themselves.
1. To conduct a lab experiment for the oscillations of a pendulum with a yielding support 2. investigate the simple pendulum
3. To prove that the distance (m) is directly proportional to period.
Theory states that T and d are related by the equation: T2 = kd3+ (4π2 l)/g where g is the acceleration of free fall and k is a constant.
Apparatus and Material:
5.Clamps holding blade
1. The length of the string holding the bob was recorded
2. The time taken for the pendulum to complete 20 oscillations was found and recorded. 3. The position of the plate was adjusted and the previous step was repeated 3 times for 6 different distances. 4. An average was found based on these 3 values.
5. Values for T2 and d3 were calculated.
The average was taken to increase accuracy = (31.50+31.47+31.44)3 = 94.413 = 31.47s The time for one oscillation was found by dividing the amount of oscillations (20) done by the time taken for them= 31.4720 = 1.57s The taking 2 points from the graph, (11.8 x 10-3, 2.45), (7 x 10-3, 2.36) Gradient= y2- y1x2- x1 = 2.45-2.361.8 x10-3-7 x10-3 = 0.090.0048 = 18.75m/s The Y intercept occurs when the X is equal to 0 and the line cuts the y- axis, hence the y- intercept is 2.23s The graph showed the relationship between both the distance and the time period. Calculations were done to determine the value of the constant k and g which is the acceleration using the length, l, of the string attached to the pendulum using the equation: T2 = kd3+4π2lg when l is 0.5m. In order to find the g, a point on the graph was substituted back into the equation to solve for it: When y = 2.23s, x= 0
T2 = kd3+4π2lg
The equation is similar to y= mx + c Where y = 2.23s, x = 0, c= 4π2lg (l = 0.5m) m= 18.75m Equation = T2 = kd3+4π2lg
solving for g= 2.23 = 18.75 (0) + 4π2(0.5)g
= 2.23 = 2π2g
=2.23g = 19.74
g = 19.742.23 = 8.85ms-1
Gravity was calculated to be 8.85ms-1
The percentage error was calculated by using the formula: % error = ±.5mm
value of d x 100. d = 0.24m 0.5mm to meters= 0.0005m % error = ± 0.00050.24 x 100
= ± 0.0021 x 100 = 0.21%
The value of k, when compared to the equation of the straight line, is equal to the gradient , m. hence k is equal to18.75m. Since the acceleration, g, was found to be 8.89ms-1 it can be seen that an error occurred whereas gravity to a free fall is said to be 9.81ms-1. As stated previously we can say that this error was caused due to the reaction time of the person recording the 20 oscillations. Question 7.1
Since T = 2secs and l = 1m
By using the formula T2 = kd3+4π2lg where T2 = 4, k=18.75, l= 1m , g= 8.85ms-1 the gradient, m, is 18.75.
Substitute (2)2 = 18.75d3+ 4π2(1)8.85
* 4 = 18.75d3 + 4.46
=4 – 4.46 = 18.75d3
= d3= -0.4618.75 = -0.025
d3 = -0.025
d = 3-0.025 = -0.29
d = 0.29m
7.2 With the length d increasing the time and angle for a period will also increase which will result in the need of more space for making its oscillations.