# Pendulum Lab Essay Sample

Abstraction

Purpose: To carry on an experiment to turn out the giving up support distance is straight relative to the period.

Consequences:

vitamin 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 giving support decreased. the period besides decreased. A cause of mistake was due to human reaction clip. Although 20 oscillations were counted. the individual keeping the stop watch might hold stopped msecs before or after. hence doing the values to be somewhat off. For this ground. an norm was taken to increase truth. This could be prevented if the individual who is numbering the oscillations besides times them themselves.

Aims

1. To carry on a lab experiment for the oscillations of a pendulum with a giving support 2. investigate the simple pendulum

3. To turn out that the distance ( m ) is straight relative to period.

Theory

Theory provinces that T and vitamin Ds are related by the equation: T2 = kd3+ ( 4?2 cubic decimeter ) /g where g is the acceleration of free autumn and K is a changeless.

Apparatus and Material:

1. Stop clock

2. Hacksaw blade

3. Stringing

4. Pendulum

5. Clamps keeping blade

6. Table

Procedure/Method

1. The length of the twine keeping the British shilling was recorded

2. The clip taken for the pendulum to finish 20 oscillations was found and recorded. 3. The place of the home base was adjusted and the old measure was repeated 3 times for 6 different distances. 4. An norm was found based on these 3 values.

5. Valuess for T2 and d3 were calculated.

Results/data

Analysis/ Discussion

The norm was taken to increase truth = ( 31. 50+31. 47+31. 44 ) 3 = 94. 413 = 31. 47s The clip for one oscillation was found by spliting the sum of oscillations ( 20 ) done by the clip taken for them= 31. 4720 = 1. 57s The taking 2 points from the graph. ( 11. 8 ten 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 clip period. Calculations were done to find the value of the changeless K and g which is the acceleration utilizing the length. l. of the twine attached to the pendulum utilizing the equation: T2 = kd3+4?2lg when cubic decimeter is 0. 5m. In order to happen the g. a point on the graph was substituted back into the equation to work out for it: When Y = 2. 23s. x= 0

T2 = kd3+4?2lg

The equation is similar to y= mx + degree Celsius Where y = 2. 23s. x = 0. c= 4?2lg ( cubic decimeter = 0. 5m ) m= 18. 75m Equation = T2 = kd3+4?2lg

work outing for g= 2. 23 = 18. 75 ( 0 ) + 4?2 ( 0. 5 ) g

= 2. 23 = 2?2g

=2. 23g = 19. 74

g = 19. 742. 23 = 8. 85ms-1

Gravity was calculated to be 8. 85ms-1

The per centum mistake was calculated by utilizing the expression: % mistake = Â± . 5mm

value of vitamin D x 100. 500 = 0. 24m 0. 5mm to meters= 0. 0005m % mistake = Â± 0. 00050. 24 ten 100

= Â± 0. 0021 ten 100 = 0. 21 %

The value of k. when compared to the equation of the consecutive 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 mistake occurred whereas gravitation to a free autumn is said to be 9. 81ms-1. As declared antecedently we can state that this mistake was caused due to the reaction clip of the individual entering the 20 oscillations. Question 7. 1

Since T = 2secs and fifty = 1m

By utilizing the expression T2 = kd3+4?2lg where T2 = 4. k=18. 75. l= 1m. g= 8. 85ms-1 the gradient. m. is 18. 75.

Substitute ( 2 ) 2 = 18. 75d3+ 4?2 ( 1 ) 8. 85

* 4 = 18. 75d3 + 4. 46

=4 â€“ 4. 46 = 18. 75d3

= d3= -0. 4618. 75 = -0. 025

d3 = -0. 025

500 = 3-0. 025 = -0. 29

vitamin D = 0. 29m

7. 2 With the length vitamin D increasing the clip and angle for a period will besides increase which will ensue in the demand of more infinite for doing its oscillations.