IB Physics Lab Report 1
Lukas Bosko 000771-008
Introduction Scales are usually used to measure the mass of a object. This means that the weight of a familiar object is compared with the weight of some unknown object with use of the static effects of the force. On the other side we are able to use dynamic effects of the force. Applying the laws of a simple harmonic oscillator we can use these dynamic effects of the force to measure mass of object.
Problem statement 1 – Is it possible to find out experimentally the spring stiffness constant? Find it out for two different springs, then stick them together and then find total spring stiffness constant. 2 – Can be the mass of an object determined using the laws of simple harmonic motion?
Hypothesis According to laws of simple harmonic motion (SHO), prolongation of spring is dependent on mass of object and spring stiffness constant. Using just one spring and changing mass of bob from trial to trial (mass is independent) can be measured prolongation(dependent variable) of a spring and from formula 1 F = k . x where F is equal to weight force and x is equal to displacement (prolongation) can be calculated spring stiffness constant which is a constant variable. In second question there are no variables manipulated. With same mechanism is measured just period of time. It means, that period (also frequency) is dependent on mass and spring stiffness constant (we use result of k from first part of experiment). Spring stiffness constant is controlled variable (held constant) and mass is dependent (real mass is also constant, but calculated mass is dependent). Measuring the period, according to T
=
2 π
m k
(formula 2) we calculate mass as m
=
T 2 .k 2
4π
(formula
3). Therefore just period (T) will be measured. Spring stiffnes constant as controlled variable is non changing.
Materials Spring, set of weights, ruler, stand, stopwatch
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IB Physics Lab Report 1
Lukas Bosko 000771-008
Procedures Springs of various stiffness constants will be used. Spring will be attached to a stand with ruler used for measuring the prolongation. The same spring will be used in one experiment. The procedure is to set up the stand with spring, attach bob to spring and measure prolongation (picture 1). For each measurement there will be another bob in question 1. From to formula (1) will be calculated spring stiffness constant.
Picture 1
After 5 measurements the spring will be changed. With another spring will be calculated its spring stiffness constant again. When both stiffness constants will be calculated (k1 and k2), springs will be connected together (picture 2). By changing mass of a bob will be measured total spring stiffness constant.
Picture 2
In second question will be spring non changing for each trial. With stopwatch will be there measured period of time (T20) for 20 cycles (from amplitude through equilibrium to another amplitude and back). That way will be calculated period of time for one cycle (T 1). Diagram of mechanism will be the same as in picture 1. From 10 measurements will be calculated mass of a bob according to formula (3). Pictures do not come out from results!
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IB Physics Lab Report 1
Lukas Bosko 000771-008
Question 1
0,07
Mass of bob
Prolongation
Spring stiffness constant
m
x
k1
(kg)
(m)
(N/m)
1.
0,06
0,023
25,59
-0,95
2.
0,08
0,031
25,32
-0,68
0,68
3.
0,10
0,040
24,53
0,11
0,11
4.
0,12
0,049
24,02
0,61
0,61
5.
0,15
0,062
23,73
0,90
0,90
k
24,64
±
0,65
l a i r T
l a i r
T
1. 2. 3. 4. 5.
k -k
0,06
| k -k| n o i t a g n o l o r P
0,95
0,05 0,04 0,03 0,02 0,01 0 0
1. 2. 3. 4. 5.
0,05
Prolongation (m)
Mass of bob
Prolongation
Spring stiffness constant
m (kg) 0,02 0,04 0,06 0,08 0,10
x (m) 0,012 0,026 0,034 0,052 0,065
k2 (N/m) 16,35 15,09 17,31 15,09 15,09
k -k
| k -k|
-0,56 0,70 -1,52 0,70 0,70
0,56 0,70 1,52 0,70 0,70
k
15,79
±
0,83
Mass of bob
Prolongation
Spring stiffness constant
m (kg) 0,02 0,04 0,06 0,08 0,10
x (m) 0,019 0,041 0,055 0,062 0,076
k (N/m) 10,33 9,57 10,70 12,66 12,91
0,91 1,66 0,53 -1,43 -1,67
0,91 1,66 0,53 1,43 1,67
k
11,23
±
1,24
k -k
| k -k|
0,1
0,15
0,2
Mass (kg)
0,07 0,06 n o i t a g n o l o r P
0,05 0,04 0,03 0,02 0,01 0 0
l a i r T
Prolongation vs. Mass
Prolongation (m)
0,02
0,04
0,06
0,08
0,1
0,12
0,02
0,04
0,06
0,08
0,1
0,12
Mass (kg)
Prolongation (m) 0,07 0,06 n o i t a g n o l o r P
0,05 0,04 0,03 0,02 0,01 0 0
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IB Physics Lab Report 1
Lukas Bosko 000771-008
Results Question 2
l a i r T
Time of 20 cyc
Time of one cyc
Mass of a bob
T20 (s) 9,8 10,3 10,2 10,5 9,9 9,8 10,1 10,3 10,4 10,0
T1 (s) 0,490 0,515 0,510 0,525 0,495 0,490 0,505 0,515 0,520 0,500
m (N/m) 0,0961 0,1062 0,1041 0,1104 0,0981 0,0961 0,1021 0,1062 0,1083 0,1001
m -m
| m -m |
0,0066 -0,0034 -0,0014 -0,0076 0,0047 0,0066 0,0007 -0,0034 -0,0055 0,0027
0,0066 0,0034 0,0014 0,0076 0,0047 0,0066 0,0007 0,0034 0,0055 0,0027
m
0,1028
±
0,004
1. 2. 3. 4. 5. 6. 7. 8. 9. 10.
m = (0.103 ±0.004) kg Used formulas:
T = 2π
m k
T 2 k
k =
F x
Mass (kg)
IB Physics Lab Report 1
Lukas Bosko 000771-008
Results Question 2
l a i r T
Time of 20 cyc
Time of one cyc
Mass of a bob
T20 (s) 9,8 10,3 10,2 10,5 9,9 9,8 10,1 10,3 10,4 10,0
T1 (s) 0,490 0,515 0,510 0,525 0,495 0,490 0,505 0,515 0,520 0,500
m (N/m) 0,0961 0,1062 0,1041 0,1104 0,0981 0,0961 0,1021 0,1062 0,1083 0,1001
m -m
| m -m |
0,0066 -0,0034 -0,0014 -0,0076 0,0047 0,0066 0,0007 -0,0034 -0,0055 0,0027
0,0066 0,0034 0,0014 0,0076 0,0047 0,0066 0,0007 0,0034 0,0055 0,0027
m
0,1028
±
0,004
1. 2. 3. 4. 5. 6. 7. 8. 9. 10.
m = (0.103 ±0.004) kg Used formulas:
m
T = 2π m=
k =
k
F x
T 2 .k 2 2
4π
5
5
∑ k i
∑ mi average m =
k =
i =1
∑ =
5
5 5
deviation
i =1
5
∑ k − k i
m − mi
i =1
deviation
5
=
i =1
5
Sample calculations 2
m=
k = k =
T .k 2 4π
F x
=
2
+
(0,49 s ) .
15,79 N .m
0,06kg.9,81ms 0,023
k 1 k 2 k 1
=
2
k 2
=
4.3,14
2
−1
=
0.0961kg
−2
=
24,64.15,79 24,64 + 15,79
25,59 N .m =
−1
389,0656 40,43
=
9,62 N .m
−1
Note: Calculations are powered by Excel (It means, that displayed part of number is different(rounded) than part used in calculation).
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IB Physics Lab Report 1
Lukas Bosko 000771-008
Conclusion and evaluation From formula (1) was calculated spring stiffness constant of spring one and spring two.
k1 = (24,64 ± 0,65) N/m k2 = (15,79 ± 0,83) N/m
Total spring stiffness constant of two connected springs can be not only measured, but also calculated by formula (4) k =
k 1 k 2 k 1
+
k 2
.
Measured value of
k = (11,23±1,24) N/m.
Calculated value of
k = 9.62 N/m
Errors are caused by condition of springs, bigger was scale of mass and prolongation, bigger was deviation. In practical conditions, it was more problematical to measure prolongation, inclusive case of two springs, where prolongation was also bigger. The springs that we used were old and their stiffness was not a constant.
In second question, the mass of a bob was calculated. Result of measurement was m=(0.103±0.004)kg what corresponds with real mass which was 0.100 kg(measured by digital scales).
Error is caused by inaccurate measurement of time because of use of cellphone stopwatch Another reason is that oscilations could be damped (it was not theoretical model of simple harmonic motion).
Limitation and weaknesses in this lab were:
-
accurate method of measuring the prolongation
-
lack of different springs
-
oscilations in Q2 was dumped, so first cycle was not equal to the last one
-
human error in measuring time of cycle exact from amplitude
Suggestion to improve this lab:
-
Using bigger stand and ruler – therefore would be there bigger increase of prelongation and could be better measured.
-
Using more accurate stopwatch for time measurement
-
Using non old damaged springs
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