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Physics 35IB
                                                                                                th
                                                                                             26 of October 2011



Background:
A simple pendulum consists of a mass called the pendulum bob suspended from a support by a thread. The
time for a complete vibration is called the period ( T ) of the pendulum. When a pendulum swings through a
small arc its bob is undergoing simple harmonic motion. This investigation attempts to verify Galileo’s early
observation that the period of a pendulum depends only upon the length of the thread, also to verify that the
acceleration due to gravity is
9.81 m / s .The concept of oscillatory motion is used. The equation related to this concept is
          2




                                                        l
                                        T = 2π
                                                        g

Problem:
The two problems that need to be solved with this Lab is to prove the dependance of the period on the length of
the rope as well as the value of gravity.

Hypothesis:
An increase in the ropes length will cause the period to grow as well. That those two variables are proportional
can be deducted in two ways. First when taking a look at the equation one can see that a multiplication of one of
the two (given that the others are constant) results in a change of the other in the same direction.

Design:
The pendulum will be set up by running a threat through a stopper which is clamped into a clamp stand. A sheet
of paper with a line indicating 20 degrees is placed some centimeters behind the string. At the bottom a
pendulum bob is tied to the string. At each trial the string is placed into the same position of 20 degrees. After
taking the time of 10 oscillations the bob is lowered and the new length is measured with a meter. This
procedure is repeated for 10 trails.




  Pendulum bob                                                             20 Degree angle
                                                                           indicated on paper
Physics 35IB
                                                                                             th
                                                                                          26 of October 2011




Variables:
Independent: The length of the rope which was increased after every trail and measured by the same meter
stick by the same person.
Corresponding: The time the pendulum bobs needs to complete 10 swings; the weight of the bob stays
constant, and so does the timer.
Controlled:
1. The stopper - the person who was in charge of the stop watch was always the same, this was to ensure that
the reaction time variations do not add additional errors to the measurement of time.
2. The same pendulum bob was used - interchanging the pendulum bob would have resulted in a different
momentum of the Pendulum for each different trial, as well as and alternation in acceleration due to the change
in mass.
3. The angle – A differentiation of the angle would have resulted in an extension/reduction of the way the bob
would swing, therefore the time needed to complete one round would have been altered.
4. The person that took the measurements - The job of measuring the length of the rope was assigned to a
single person. This way, adding unnecessary errors while taking the length was avoided as there was just one
value of uncertainty, that of the analyzer.
5. The environment – The set up was not moved in between the trails so the environment was kept more or less
the same from trail to trail. Factors like wind, pressure or temperature did not change greatly since the
apparatus was not displaced. Neither was any window or door opened which could have caused wind gushes to
influence the results.


Materials:
   • A pendulum bob with a certain mass
   • Pendulum clamps
   • Stop watch
   • String
   • Paper with 20 degree angle indicated
   • Ruler
   • Stopper with hole



Procedure:
   1. Attach String to Pendulum clamps and run it through the hole of the Stopper
   2. Tie Pendulum bob to lose end of String
   3. Measure and record the length of the rope
   4. Align String so it is in line with the indicated 20 degree angle
   5. Release Pendulum and start the Stop watch.
   6. Count 10 oscillations and record the time
   7. Modify length and repeat steps 4.-6. until 10 trails have been performed
Physics 35IB
                                                                                                                      th
                                                                                                              26 of October 2011



In order to attain a linear graph from the first parabolic graph created using the data from table 1.0. the
Period must to be squared. This can be deduced from the original equation                 l   where we
                                                                                                    T = 2π
find the length (l ) under a square root.                                                                     g
The equation for the first graph can be obtained by replacing square rooted l the of the original
equation by   l as shown here:                    1         The gradient for the first graph is                   1
                                       T = 2π       *l                                                   2π
                                                  g                                                               g


    Calculations for T2:
    T = period
    T2 = (period)2
    ex. Data from trial 1
    T2 = (1.035 s)2
    T2 = 1.071 s2

          Length ± 0.001            Period2 ± 2%
               (m)                        (s2)
                 0.276                    1.071
                 0.325                    1.201
                 0.386                    1.543
                 0.439                    1.703
                 0.479                    1.869
                 0.518                    1.999
                 0.557                    2.135
                 0.646                    2.605
                 0.754                    2.938
                 0.824                    3.349
          Table 1.1 – Period squared as a function of the length of the rope


Uncertainties:
The uncertainty of the length 0.001m (1cm) was set by the person who measured the rope. The max.
uncertainty of 0.0005m could not be reached because of the combination of her bad eye sight and the
faded unit indications of the ruler.
The beforehand measured reaction time of the stopper (0.15s) was used as uncertainty with the 'Time
of 10 oscillations'.
The error for the actual Period was calculated in the following manner:
       1
 ∆P =    ∆ t t = time for ten oscillations
      10


In order to define the error of T2 another calculation must be made.
The formula needed to receive the error of a power equation is like z=xn
                                                                                                     ∆T 
                                                                                 ∆T   2
                                                                                          = T   2
                                                                                                    2   
                                                                                                      T 
Physics 35IB
                                                                                                                              th
                                                                                                                           26 of October 2011



        ∆ x  After in to manipulating the formula to fit the needs:
∆z = z  n   
          x 

When replacing the variables with the real values one finds that the values vary from Data Point to
Data Point. So to solve the problem one can, instead of taking the exact values of uncertainty,
calculate the percent uncertainty. The equation needs to be modified slightly

                                                      2
                                                           ∆T 
                                                 T        2      
                                                               T 
               P e r c e n ta g e ∆T     2
                                             =                      •1 0 0 %
                                                            T 2
ex. using Data from 7th trail
                                                 0 .0 1 5 s                                                       0 .0 4 3 8 4
                                  2 .1 3 5 s 2  2                           P e r c e n ta g e ∆ 2 .1 3 5 s 2 =                •1 0 0 %
                                                 1 .4 6 1 s  • 1 0 0 %                                            2 .1 3 5 s 2
P e r c e n ta g e ∆ 2 .1 3 5 s =
                               2

                                            2 .1 3 5 s 2                      P e r c e n ta g e ∆ 2 .1 3 5 s 2 = 2 %

After computing all of the percent uncertainty one comes to the conclusion that the average
percentage error is 2%.

Error Bars:
Error Bars are assigned to the higher Percent Uncertainty. From the calculations above one can see
the error of Period squared is 2%.


Percent Uncertainty of length is 0.4%, this result can be attained from the following process:
                           0 .0 0 1 m
P e r c e n ta g e ∆ l =
                           0 .2 7 6 m
                                      •100%          P e r c e n ta g e ∆ l = 0 .4 %
Two percent is the greater value here and therefore only vertical Error Bars are used in the graph.

Maximum and Minimum Line of Best Fit:
Data Points for Max. LofBF:
Lowest Point = xmin, ymin - error(ymin)                                         Highest Point = xmax, ymax + error(ymax)
               0.276m, 1.071s2 – 2%x1.071s2                                             0.824m, 3.349s2 + 2%x3.349s2
              (0.276m, 1.040s2)                                                        (0.824m, 3.404s2)


Data Points for Min. LofBF:
Lowest Point: xmin, ymin + error(ymin)
                0.276m, 1.071s2 + 2%x1.071s2
               (0.276m, 1.102s2)
Physics 35IB
                                                                                        th
                                                                                      26 of October 2011



Highest Point = xmax, ymax - error(ymax)
                     0.824m, 3.349s2 - 2%x3.349s2
                     (0.824m, 3.294s2)
Equations of Graphs:
Using two points of the graph one can calculate the slope of a linear graph with the following equation:
            y − y1
 s lo p e = 2
            x 2 − x1




Line of Best Fit Slope:
When using the points of the second(0.325m, 1.201s 2) and seventh (0.557m,2.135s2)
trail         2 .1 3 5 s 2 − 1 .2 0 1 s 2 s2 the value of the slope is 4.026s2m-1
        s lo p e =                                  = 4 .0 2 6
                      0 .5 5 7 m − 0 .3 2 5 m                    m
The slope on the graph varies from the one just calculated since the value of 4.026s 2m-1 is limited due
to the selection of the 2nd and 7th point.
Equation: 4.107s2m-1x – 0.096

Minimum Line of Best Fit:
Using the points (0.276m, 1.102s2) and (0.824m, 3.249s2) and following the same steps as above
             3 .2 4 9 s 2 − 1 .1 0 2 s 2            s2
s lo p e =                               = 3 .9 1 7
             0 .8 2 4 m − 0 .2 7 6 m                m
Equation: 3.061s2m-1x+0.526



Maximum Line of Best Fit:
Using the points (0.276m, 1.040s2) and (0.824m, 3.404s2) and following the same steps as above
             3 .4 0 4 s 2 − 1 .0 4 0 s 2            s2
s lo p e =                               = 4 .3 1 4
             0 .8 2 4 m − 0 .2 7 6 m                m
Equation: 5.251s2m-1x-0.678



Uncertainty for Slope Calculations:
Physics 35IB
                                                                                                                                                  th
                                                                                                                                                26 of October 2011



∆ s lo p e = S lo p e o f M a x im u m L o fB F − S lo p e o f M in im u m L o fB F
                                 −1                          −1
∆ s lo p e = 4 .3 1 4 s 2 m           − 3 .9 1 7 s 2 m
                                −1
∆ s lo p e = 0 .3 9 7 s 2 m
                  1                                               l                                    l
Slope = 4     π                Based on: T = 2 π                      → T
                                                                                  2
                                                                                      = 4π         2

                  g                                               g                                    g




Acceleration due to Gravity:
After defining the slope it is now possible to solve for the last unknown variable g of the original
equation
           T = 2π
                   l                   l .                                                l                                 1                              1m
                           g
                               ( )2      T 2 = 4π        2
                                                                       g =                        * 4π     2
                                                                                                                   g =            * 4π      2
                                                                                                                                                 g =                * 4π   2

                                                             g                        T       2
                                                                                                                         S lo p e                      4 .0 2 6 s 2
                                                                          l                  1
                                                                              2
                                                                                  =
                      m                                               T                   S lo p e
g = 9 .6 1 2
                      s2
The uncertainty of Gravity:
Gravity = 9.612ms-2 ± 0.945ms-2


              ∆ S lo p m   0 .3 9 7 m s 
                                              2
                           e                                                                                                           m
 ∆∆g g = =g 9 . 6 1 2 2÷                    2 ÷                                                                     ∆ g = 0 .9 4 5
                 S l o p es   4 . 0 2 6 m s                                                                                        s2

Comparison to the actual acceleration due to gravity value(9.81ms -2):

                                                                                                                  m            m
                                                                                                       9 .6 1 2       − 9 .8 1 2
                           e x p e r im e n ta l − th e o r e tic a l                                             s 2
                                                                                                                              s •10 0%
% D iffe r e n c e =                                                  •100% =
                                      e x p e r im e n ta l                                                              m
                                                                                                               9 .6 1 2 2
                                                                                                                        s
% D iffe r e n c e = 2 %
Physics 35IB
                                                                                       th
                                                                                    26 of October 2011



Evaluation & Conclusion:

Conclusion:




Evaluation:
After calculating the percent difference between the accepted value of gravity(9.81ms -2) and the
experimental(9.612ms-2) with the result of 2% it is discernible that the experiment was completed with
great accuracy. The difference of 2% is uncommonly low however there were still some factors which
during the experiment have affected and altered the values from the 'perfect' ones which would have
resulted in a difference of 0%.

Some of the errors that affected the experiment are:

1. The stability of the clamp stand or to be


   1. Unstable Clamp stand. It wobbles at the first swing and it follows the
      vibration of the bob. Therefore it adds an extra acceleration to the bob.
   2. The stopper which the string runs through has an circular hole. A lot of
      space is available for derivation from a straight line oscillation. The
      circular hole even benefits a non-linear motion.
   3. The angle differs from trail to trail.
   4. The person letting the bob go and the time stopper do not start at the same
      time.
   5. Person letting the bob go might unconsciously add force.

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IB Phyiscs SL - Pendulum Lab

  • 1. Physics 35IB th 26 of October 2011 Background: A simple pendulum consists of a mass called the pendulum bob suspended from a support by a thread. The time for a complete vibration is called the period ( T ) of the pendulum. When a pendulum swings through a small arc its bob is undergoing simple harmonic motion. This investigation attempts to verify Galileo’s early observation that the period of a pendulum depends only upon the length of the thread, also to verify that the acceleration due to gravity is 9.81 m / s .The concept of oscillatory motion is used. The equation related to this concept is 2 l T = 2π g Problem: The two problems that need to be solved with this Lab is to prove the dependance of the period on the length of the rope as well as the value of gravity. Hypothesis: An increase in the ropes length will cause the period to grow as well. That those two variables are proportional can be deducted in two ways. First when taking a look at the equation one can see that a multiplication of one of the two (given that the others are constant) results in a change of the other in the same direction. Design: The pendulum will be set up by running a threat through a stopper which is clamped into a clamp stand. A sheet of paper with a line indicating 20 degrees is placed some centimeters behind the string. At the bottom a pendulum bob is tied to the string. At each trial the string is placed into the same position of 20 degrees. After taking the time of 10 oscillations the bob is lowered and the new length is measured with a meter. This procedure is repeated for 10 trails. Pendulum bob 20 Degree angle indicated on paper
  • 2. Physics 35IB th 26 of October 2011 Variables: Independent: The length of the rope which was increased after every trail and measured by the same meter stick by the same person. Corresponding: The time the pendulum bobs needs to complete 10 swings; the weight of the bob stays constant, and so does the timer. Controlled: 1. The stopper - the person who was in charge of the stop watch was always the same, this was to ensure that the reaction time variations do not add additional errors to the measurement of time. 2. The same pendulum bob was used - interchanging the pendulum bob would have resulted in a different momentum of the Pendulum for each different trial, as well as and alternation in acceleration due to the change in mass. 3. The angle – A differentiation of the angle would have resulted in an extension/reduction of the way the bob would swing, therefore the time needed to complete one round would have been altered. 4. The person that took the measurements - The job of measuring the length of the rope was assigned to a single person. This way, adding unnecessary errors while taking the length was avoided as there was just one value of uncertainty, that of the analyzer. 5. The environment – The set up was not moved in between the trails so the environment was kept more or less the same from trail to trail. Factors like wind, pressure or temperature did not change greatly since the apparatus was not displaced. Neither was any window or door opened which could have caused wind gushes to influence the results. Materials: • A pendulum bob with a certain mass • Pendulum clamps • Stop watch • String • Paper with 20 degree angle indicated • Ruler • Stopper with hole Procedure: 1. Attach String to Pendulum clamps and run it through the hole of the Stopper 2. Tie Pendulum bob to lose end of String 3. Measure and record the length of the rope 4. Align String so it is in line with the indicated 20 degree angle 5. Release Pendulum and start the Stop watch. 6. Count 10 oscillations and record the time 7. Modify length and repeat steps 4.-6. until 10 trails have been performed
  • 3. Physics 35IB th 26 of October 2011 In order to attain a linear graph from the first parabolic graph created using the data from table 1.0. the Period must to be squared. This can be deduced from the original equation l where we T = 2π find the length (l ) under a square root. g The equation for the first graph can be obtained by replacing square rooted l the of the original equation by l as shown here: 1 The gradient for the first graph is 1 T = 2π *l 2π g g Calculations for T2: T = period T2 = (period)2 ex. Data from trial 1 T2 = (1.035 s)2 T2 = 1.071 s2 Length ± 0.001 Period2 ± 2% (m) (s2) 0.276 1.071 0.325 1.201 0.386 1.543 0.439 1.703 0.479 1.869 0.518 1.999 0.557 2.135 0.646 2.605 0.754 2.938 0.824 3.349 Table 1.1 – Period squared as a function of the length of the rope Uncertainties: The uncertainty of the length 0.001m (1cm) was set by the person who measured the rope. The max. uncertainty of 0.0005m could not be reached because of the combination of her bad eye sight and the faded unit indications of the ruler. The beforehand measured reaction time of the stopper (0.15s) was used as uncertainty with the 'Time of 10 oscillations'. The error for the actual Period was calculated in the following manner: 1 ∆P = ∆ t t = time for ten oscillations 10 In order to define the error of T2 another calculation must be made. The formula needed to receive the error of a power equation is like z=xn  ∆T  ∆T 2 = T 2 2   T 
  • 4. Physics 35IB th 26 of October 2011  ∆ x  After in to manipulating the formula to fit the needs: ∆z = z  n   x  When replacing the variables with the real values one finds that the values vary from Data Point to Data Point. So to solve the problem one can, instead of taking the exact values of uncertainty, calculate the percent uncertainty. The equation needs to be modified slightly 2  ∆T  T 2   T  P e r c e n ta g e ∆T 2 = •1 0 0 % T 2 ex. using Data from 7th trail  0 .0 1 5 s  0 .0 4 3 8 4 2 .1 3 5 s 2  2  P e r c e n ta g e ∆ 2 .1 3 5 s 2 = •1 0 0 %  1 .4 6 1 s  • 1 0 0 % 2 .1 3 5 s 2 P e r c e n ta g e ∆ 2 .1 3 5 s = 2 2 .1 3 5 s 2 P e r c e n ta g e ∆ 2 .1 3 5 s 2 = 2 % After computing all of the percent uncertainty one comes to the conclusion that the average percentage error is 2%. Error Bars: Error Bars are assigned to the higher Percent Uncertainty. From the calculations above one can see the error of Period squared is 2%. Percent Uncertainty of length is 0.4%, this result can be attained from the following process: 0 .0 0 1 m P e r c e n ta g e ∆ l = 0 .2 7 6 m •100% P e r c e n ta g e ∆ l = 0 .4 % Two percent is the greater value here and therefore only vertical Error Bars are used in the graph. Maximum and Minimum Line of Best Fit: Data Points for Max. LofBF: Lowest Point = xmin, ymin - error(ymin) Highest Point = xmax, ymax + error(ymax) 0.276m, 1.071s2 – 2%x1.071s2 0.824m, 3.349s2 + 2%x3.349s2 (0.276m, 1.040s2) (0.824m, 3.404s2) Data Points for Min. LofBF: Lowest Point: xmin, ymin + error(ymin) 0.276m, 1.071s2 + 2%x1.071s2 (0.276m, 1.102s2)
  • 5. Physics 35IB th 26 of October 2011 Highest Point = xmax, ymax - error(ymax) 0.824m, 3.349s2 - 2%x3.349s2 (0.824m, 3.294s2) Equations of Graphs: Using two points of the graph one can calculate the slope of a linear graph with the following equation: y − y1 s lo p e = 2 x 2 − x1 Line of Best Fit Slope: When using the points of the second(0.325m, 1.201s 2) and seventh (0.557m,2.135s2) trail 2 .1 3 5 s 2 − 1 .2 0 1 s 2 s2 the value of the slope is 4.026s2m-1 s lo p e = = 4 .0 2 6 0 .5 5 7 m − 0 .3 2 5 m m The slope on the graph varies from the one just calculated since the value of 4.026s 2m-1 is limited due to the selection of the 2nd and 7th point. Equation: 4.107s2m-1x – 0.096 Minimum Line of Best Fit: Using the points (0.276m, 1.102s2) and (0.824m, 3.249s2) and following the same steps as above 3 .2 4 9 s 2 − 1 .1 0 2 s 2 s2 s lo p e = = 3 .9 1 7 0 .8 2 4 m − 0 .2 7 6 m m Equation: 3.061s2m-1x+0.526 Maximum Line of Best Fit: Using the points (0.276m, 1.040s2) and (0.824m, 3.404s2) and following the same steps as above 3 .4 0 4 s 2 − 1 .0 4 0 s 2 s2 s lo p e = = 4 .3 1 4 0 .8 2 4 m − 0 .2 7 6 m m Equation: 5.251s2m-1x-0.678 Uncertainty for Slope Calculations:
  • 6. Physics 35IB th 26 of October 2011 ∆ s lo p e = S lo p e o f M a x im u m L o fB F − S lo p e o f M in im u m L o fB F −1 −1 ∆ s lo p e = 4 .3 1 4 s 2 m − 3 .9 1 7 s 2 m −1 ∆ s lo p e = 0 .3 9 7 s 2 m 1 l l Slope = 4 π Based on: T = 2 π → T 2 = 4π 2 g g g Acceleration due to Gravity: After defining the slope it is now possible to solve for the last unknown variable g of the original equation T = 2π l l . l 1 1m g ( )2 T 2 = 4π 2 g = * 4π 2 g = * 4π 2 g = * 4π 2 g T 2 S lo p e 4 .0 2 6 s 2 l 1 2 = m T S lo p e g = 9 .6 1 2 s2 The uncertainty of Gravity: Gravity = 9.612ms-2 ± 0.945ms-2  ∆ S lo p m   0 .3 9 7 m s  2 e m ∆∆g g = =g 9 . 6 1 2 2÷  2 ÷ ∆ g = 0 .9 4 5  S l o p es   4 . 0 2 6 m s  s2 Comparison to the actual acceleration due to gravity value(9.81ms -2): m m 9 .6 1 2 − 9 .8 1 2 e x p e r im e n ta l − th e o r e tic a l s 2 s •10 0% % D iffe r e n c e = •100% = e x p e r im e n ta l m 9 .6 1 2 2 s % D iffe r e n c e = 2 %
  • 7. Physics 35IB th 26 of October 2011 Evaluation & Conclusion: Conclusion: Evaluation: After calculating the percent difference between the accepted value of gravity(9.81ms -2) and the experimental(9.612ms-2) with the result of 2% it is discernible that the experiment was completed with great accuracy. The difference of 2% is uncommonly low however there were still some factors which during the experiment have affected and altered the values from the 'perfect' ones which would have resulted in a difference of 0%. Some of the errors that affected the experiment are: 1. The stability of the clamp stand or to be 1. Unstable Clamp stand. It wobbles at the first swing and it follows the vibration of the bob. Therefore it adds an extra acceleration to the bob. 2. The stopper which the string runs through has an circular hole. A lot of space is available for derivation from a straight line oscillation. The circular hole even benefits a non-linear motion. 3. The angle differs from trail to trail. 4. The person letting the bob go and the time stopper do not start at the same time. 5. Person letting the bob go might unconsciously add force.