The association of Hooke’s law with elastic were investigated by hanging masses off an elastic cord, and measuring the extensions. A graph of the results was made to determine Hooke’s constant which would later be applied to certain calculations. Then a model bungee jump was created to measure the motions and energies involved with bungee jumping.

From the results gathered, calculations and predictions were made in order to provide the foundation of the construction, and properties of real bungee cords which could be applied when jumping off the Story Bridge.Aim: To investigate the factors and properties associated with a bungee cord, in order to design a suitable model bungee jump, which will be applied as the basis for an actual bungee jump attraction on the Story Bridge.Introduction and Background Information:This investigation into bungee jumping was initiated as a result of a company’s proposal to construct a bungee jump attraction on the Story Bridge. Bungee jumping is a popular extreme sport which originated from the Pentecost Islands of Vanuatu, where the men tied vines around their ankles and jumped from a height as a test of courage. Today bungee jumping consists of an elastic cord secured to a platform and a variety of other equipment. Modern bungee jumping, with an elastic cord, demonstrates Hooke’s Law and consists of a variety of energy transformations. According to Hooke’s law the spring constant is theoretically linear, and is calculated by k=Fï¿½, therefore is the gradient of the line which passes through the origin.

There are many important variables to consider when designing a bungee jump such as Hooke’s (spring) constant, mass, weight, length of the chord, the length of extension, height, force, velocity, acceleration, gravitational potential energy, elastic potential energy and kinetic energy. Each of these factors contributes to, and affects different aspects of the bungee jump, but together can impact the overall effectiveness.When applying the model to real life it is important to know the Hooke’s constant as further predictions and calculations can then be applied to determine other variables such as the appropriate length of the chord, the extension and force. Length, extension and force are extremely crucial because if the person is too heavy the elastic will extend too much, if the person is too light the jump will be jolty and possibly cause injury.

Therefore extensive planning and research has to be conducted in order to design an effective model, which could then be applied as the foundation for the Story Bridge bungee jump. Hooke’s Law is the relationship between the force exerted on a spring or elastic and the extension.Apparatus:* elastic* Masses 50-500g* Mass holder* Stand* Ruler* Plastic Container Lid* Tripod* Glx motion sensor on person mode* Data Studio* Bench* BlockProcedure:Two experiments were conducted in order to find basic measurements and data involved in calculations. The first experiment was carried out to find the extensions of the elastic for various amounts of mass, with the purpose of determining Hooke’s (spring) constant. Firstly, a strand of elastic was halved and tied together to produce a chord with two parallel elastics.

The elastic was then secured onto the pole extension of the stand, and the length of the resulting elastic was then measured. The stand was placed on a bench with the string hung over the ledge. Masses of 50g to 500g were placed onto the mass holder, which was also measured, and then hung on the suspended string. The length between the top of the string and the bottom of the mass holder was measured for each new mass. From the recorded data, the extensions of the string was calculated and a force-extension graph was produced in order to calculate Hooke’s constant, which was the rate of change in force needed to extend the string per metre.After the findings of the first experiment were recorded, the second experiment was carried out using the same elastic to record the motions of a bungee jump. A mass of 300g was attached to the elastic cord and secured with blu tac, and the motion sensor was put on the ground directly underneath the suspended string. The motion sensor was placed on a block under a tripod for equipment safety, and the block was utilised to ensure that there was no interference between the tripod and the sensor.

A plastic container lid was secured with blu tac on the bottom of the masses to maximise the surface area and prevent other objects interfering with the sensor. Blu tac was positioned on top of the masses, to secure the masses during the fall. After the equipment was made secure, the 300g mass was brought up to the pole extension, where the string was secured. The motion sensor was initiated and the 300g mass dropped. A position-time graph was obtained from the GLX motion sensor using Data Studio.Results:EXPERIMENT 1- Hooke’s ConstantLength of folded string (2 strings parallel): 42.

8 cmLength of mass holder: 11.9cmExtension of the string: measured length (extended string + mass holder) – Original length of stringMass(kg)Force(N)Measured Length(cm)Extension(cm)0042.800.050.4954.700.1000.9854.

90.20.1501.

4755.40.70.2001.9656.

61.90.2502.45594.30.

3002.9463.18.40.3503.436813.

30.4003.9273.

118.40.4504.4178.223.50.5004.982.

427.7Graph of dataEXPERIMENT 2Position-time graph of a bungee jump for the mass of 0.3kg (2.94N)Maximum Extension:Original length of string (2 strings parallel): 42.8 cmLength of mass holder: 11.

9cmMeasured length : 94cmExtension of the string: measured length (extended string + mass holder) – Original length of stringExtension: 39.3cmLength of extended cord: 82.1cmConstant: 9.74Nmï¿½ï¿½Analysis and Discussion:According to the general theory of Hooke’s law, for every metre the elastic extends there is a constant rate of change in force. As a result the graph should be linear and the equation to determine the constant k=Fï¿½. However, when experiment 1 was conducted to establish the Hooke’s constant of two cords in parallel, the data when graphed progressed into a partially linear graph but was not completely linear. At 0g to 50g there was no extension and from above 50g to 250g the elastic extended slightly, but in a non linear manner.

This is explainable as for different materials of elastic, certain masses would have little or no effect on the extension, and the elastic would only extend at a constant rate for certain mass and onwards. Between 300g to 450g the extension appears to change at a constant rate and is dominantly linear, but at 500g the constant changes and alters the linear trend. It is assumed that once the elastic develops into a linear rate of change, it should continue to do so unless it reaches its elastic limit. For this reason there was most likely an error in measurement for the extension of elastic for 500g.

With correct measurements the graph would continue in a dominantly linear trend. Because the graph was not linear, an average rate of change was calculated from the points at which the rate of change was predominantly constant. This was for the masses 300g to 450g and the average constant was 9.74Nmï¿½ï¿½.There is a probability that the results were inaccurate.

The results were derived from the use of a ruler, so there may have been minor miscalculations of a minimum 0.5mm. In addition the assumption was made that the masses were according to what was labelled on each mass, and was not measured to certify that each mass was accurate to their label. However, if the masses labelled were inaccurate, it would probably have little impact or no relevancy as it would most likely be incorrect by a minor amount.From this graph a general equation was formed to predict the force required for the elastic to extend by . The formula was obtained by creating a linear y=m+ c equation where y was F(force), m was K(constant). The equation is F=k+ c and not F=k because the line does not go through the origin. This equation is only effective for masses of 300g and above, as smaller masses have little impact on the spring and do not extend at a constant rate.

This equation determines the minimum extensions and the forces required.F=k+ cF= 2.94NK= 9.

74Nmï¿½ï¿½= 8.4cmï¿½0.084mc=?c= F -kc= 2.94- 9.74 ï¿½ 0.

084= 2.94- 0.818= 2.123F= 9.

74 +2.123The results from experiment 2 were gathered from a position time graph. The graph displays the motion of the bungee cord and the 300g mass oscillating up and down, with each cycle gradually decreasing. Ideally, what extends down must extend back up. Realistically, friction and other forces acting upon the bungee prevent this from happening, and therefore the amount the bungee extends downwards also decreases. The maximum extension of the elastic for 300g is 39.3 cm, but eventually the will rest at its minimum extension, which was calculated in experiment 1 as 8.

4cm.Experiment 2 was performed with a mass of 300g in order to evaluate the energies and the movement involved in a bungee jump. A position-time graph was recorded from the motion sensor, and calculations from this data were developed. These calculations were based on the results obtained from the graph, such as the maximum extension of the bungee cord. Since the results were obtained from the graph, it is extremely likely that there would have been errors in recorded measurements, as the results were predominantly approximated.

However if there was any error it would only be incorrect by approximately 2cm, which could have a minor effect when calculating and predicting the accurate extensions and energies experienced on a real bungee. The mass of the blu tac and container lid, which may have had a slight effect on the results, were not taken into account. Interference between the motion sensor and other objects may have occurred and would have been recorded on the graph.From experiment 1 and 2:From experiment 1, it was found that the minimum extension of a mass of 300g was 8.4cm. To find out the maximum extension of the elastic cord for the mass of 300g, the mass was dropped from a height.

The maximum extension for that cord at that particular height was measured to be 39.3cm.Experiment 1At 2.94 N (0.3kg) the string extends by 8.4cmExperiment 2At 2.

94 N (0.3kg) the string extends by 39.3cmSince in both experiments the same string was used, according to Hooke’s law the constant must be the same. However the same force has been applied but yet the extensions are different. This can only be explained by the differences in potential energies which are affected by displacement. Experiment 2 was trialled with the masses falling from a height, whereas in experiment 1 the masses were simply hung.If experiment 2 was to be continued with different masses, assuming that the masses fall from the same place and given that the constants are the same, when drawing a force- extension graph, it should be linear and parallel to that of the extensions in experiment 1.

Therefore the general linear equation can also be applied to experiment 2, and can determine the maximum extensions of the string when dropped.Max extension (effective for masses 300g and above):F=k + cF=2.94NK=9.74Nmï¿½ï¿½=39.3cmï¿½0.

393mc=?c= F -kc= 2.94- 9.74ï¿½ 0.393=2.

94- 3.827= -0.887F=9.74 -0.887From this information it is possible to calculate measurements which can be applied to real life.

Measurements of the Story Bridge which are relevant to this investigation were researched. The jumper must jump from the pathway or an extension coming off of the pathway, as jumping off the summit would result in injury due to collision with, or entanglement on the structure. This means that the jump will not be extremely high and the clearance between the pathway and the water would have to be considered in order to calculate significant measurements.

Story Bridge proposed measurements:Maximum clearance of bridge to water: 35mMaximum height of person: 2mPerson to water: 20cmMaximum length of extended cord: 32.8mExperiment measurements:Original length of string (2 strings folded): 42.8 cmï¿½ 0.428Max Extension: 39.

3cmï¿½0.393mMax length of extended cord: 82.1cmï¿½0.821mConstant: 9.74Nmï¿½ï¿½The number of extended experiment cords needed in a series to produce the length of the extended Story Bridge cord is calculated by dividing the preferred maximum length of the bungee cord by the maximum length of the actual cord. From calculating this, the actual length of the cord and maximum extension can be calculated.The number of extended experiment cords needed in a series to produce the length of the extended Story Bridge cord:32.8ï¿½0.

821=39.95ï¿½ to be more convenient round up to 40 cordsActual length of cord (length of string ï¿½ actual number of cords):0.428ï¿½40= 17.12mMax extension of actual cord:0.

393ï¿½40=15.72mMinimum extension:0.084ï¿½40=3.36mLength of maximum extended cord:17.12+15.

72= 32.84mLength of minimum extended cord:17.12+3.36=20.48mUsing the general equation for the maximum extension of the elastic cord, the exerted force needed to extend the cord by 15.72m can be calculated.F=9.74 -0.

887F=?=15.72mF=9.74ï¿½15.72- 0.985=153.112-0.

985=152.127Nï¿½15.52kgTo accommodate a wide range of clients it is more efficient to have two bungee cords. One which supports a maximum of approximately 90kg and the other which holds a maximum of approximately 150kg.So far it has been calculated:Story Bridge measurements:Actual length of cord (2 parallel): 17.12mMax extension of cord: 15.72mLength of extended cord: 32.84mMax mass: 15.

52kg ï¿½152.523NConstant: 9.74Nmï¿½ï¿½Bungee cord 1Length: 17.12mMax extension of cord: 15.72mLength of extended cord: 32.84mMax mass: 90kg ï¿½882NConstant:?If the strands of elastic in parallel are the same, each individual strand of elastic will have the same constant, which contributes to the greater overall constant of the parallel strands working together. For example if a strand of elastic had a constant of 5 Nmï¿½ï¿½ a cord which consisted of four of these strands would have a constant of 5 + 5 + 5 + 5, or in other words the number of strands multiplied by the constant 4ï¿½5=20 Nmï¿½ï¿½.

If the strands are in parallel and a force is applied, each strand will share the force equally. So if there is eight strands in parallel and a 40N force applied, each individual strand will support 5N.Each string has the same constant:Therefore if 2 strings have a constant of 9.74Nmï¿½ï¿½One string must have half of that, so the constant of one string is 4.87Nmï¿½ï¿½If the cords are in parallel they share the force equally:Therefore if 2 strings extend 15.72m with a force of 152.

523NOne string would extend 15.72m with a force 76.06NNumber of chords needed to support 882N (90kg):882ï¿½76.06= 11.

59ï¿½for convenience use 12 cords in parallelThe actual mass supported by 12 cords in parallel:12ï¿½76.06= 912.72Nï¿½93.

135kgConstant of 12 strings:12ï¿½4.87= 58.44 Nmï¿½ï¿½Length (12 cords parallel): 17.12mMax extension of cord: 15.72mLength of extended cord: 32.84mMax mass: 93.135kg ï¿½912.72NConstant: 58.

44 Nmï¿½ï¿½The general equation for the maximum extensions of this cord would be:F=k + cF=912.72NK=58.44Nmï¿½ï¿½=15.72mc=?c= F -kc= 912.72- 58.

44ï¿½ 15.72= 912.72- 918.67= -5.95F=58.

44-5.95For safety precautions, the company prefers to have a minimum mass required. This minimum mass should have a maximum extension of at least 5m, to reduce the risk of injury from jolting.Minimum mass/weight requirement:F=58.44-5.95=5mF=58.

44ï¿½5-5.95=286.25Nï¿½29.21kgFor two parallel strands the equation is only effective for 300g and above. Therefore for one string it must be 150g and above. To calculate the minimum mass requirement for which the equation F=58.

44-5.95 starts to be effective:150ï¿½12= 1800gï¿½1.8kgï¿½17.

64NUsing similar steps a design for bungee cord 2 was created.Bungee Cord 2Length: 17.12mMax extension of cord: 15.72mLength of extended cord: 32.84mMax mass: 150kgï¿½1470NConstant: ?Number of chords needed to support 1470N (150kg):1470ï¿½76.

06= 19.32ï¿½for convenience use 20 cords in parallelActual Mass supported by 20 cords in parallel:20ï¿½76.06= 1521.2Nï¿½155.22kgConstant of 20 strings:20ï¿½4.87=97.4 Nmï¿½ï¿½Length (20 cords parallel): 17.12mMax extension of cord: 15.

72mLength of extended cord: 32.84mMax mass: 155.22kg ï¿½1521.

2NConstant: 97.4 Nmï¿½ï¿½The general linear equation for the maximum extensions of this cord would be:F=k + cF=1521.2NK=97.4Nmï¿½ï¿½=15.72mc=?c= F -kc= 1521.2- 97.4ï¿½ 15.

72= 1521.2- 1531.128= -9.928F=97.4 – 9.928Minimum mass/ weight requirement:F=97.4 – 9.

928=5mF=97.4ï¿½5-9.928=477.072N48.68kgHowever Bungee cord 1 will expand more for this weight, so it would be common sense to use bungee cord 1. It is more appropriate to use bungee cord 2 from 912.

72N (maximum weight of bungee cord 1) and onwards.To calculate the minimum mass requirement for which the equation F=97.4 – 9.928 starts to be effective:150ï¿½20=3000gï¿½3kgï¿½29.

4NMaximum extension of cord 2 with a weight of 912.72NF=k + cF=97.4 – 9.

928= (F-c) ï¿½kF=912.72Nk=97.4Nmï¿½ï¿½c=-9.928= (912.72+9.928)ï¿½97.4= 922.

648 ï¿½97.4=9.47mEnergiesA bungee jump is an excellent example of energy forms and transformations. Bungee jumps have a great amount of potential energy, which is the potential to move, and is dictated by position. Bungee jumps also have high energy due to motion, which is kinetic energy. At different stages of a jump some energies are present or will be later converted to a different form of energy.

At the very start of the jump, there is maximum gravitational potential energy, but no kinetic energy or elastic potential energy. During free fall, when the jumper experiences the acceleration of gravity, gravitational potential energy starts to decrease, but kinetic energy increases. At the very last moment of free fall, kinetic energy is at its maximum, which also means that the velocity is highest at this point. When the cord starts to extend, elastic potential energy is gained and starts to increase. The velocity at this point decreases, and therefore kinetic energy also decreases.

At the maximum extension of the jump there is zero velocity, zero kinetic energy, zero gravitational potential energy and the force down is the same as the force up. However at this point elastic potential energy is at its peak. In theory at this point gravitational has been all converted into elastic potential energy, but as a result of wind resistance, conversion of energy into heat and noise and other factors, a substantial amount of energy will be lost.

These trends are shown from the calculations below:The energies experienced on a bungee jump were calculatedExperiment 2:At the start of the jumpGPE=mghm=0.3kgg=9.8msï¿½ï¿½h=0.821mGPE=0.3ï¿½9.8ï¿½0.821=2.

414 JEPE=0KE=0Last moment of Freefallvï¿½=uï¿½ + 2asu=0msï¿½ï¿½a=9.8msï¿½ï¿½s=0.428mv==2.

896 msï¿½ï¿½v=u+att= (v-u)ï¿½a=2.896ï¿½9.8=0.295sKE=1/2mvï¿½m=0.3kgv=2.896 msï¿½ï¿½KE=1/2ï¿½0.3ï¿½2.

896ï¿½=1.258 JGPE=mghm=0.3kgg=9.8msï¿½ï¿½h=0.393mGPE=0.3ï¿½9.8ï¿½0.393=1.

155 JEPE=0At maximum extensionEPE=1/2kï¿½k=9.74Nmï¿½ï¿½=0.393mEPE=1/2ï¿½9.74ï¿½0.393ï¿½=0.752 JForceF=-(k+c)F=9.74-0.887k=9.

74Nmï¿½ï¿½=0.393mF=-(9.74ï¿½0.393-0.887)=-2.94NGPE=0KE=0At minimum extensionEPE=1/2kï¿½k=9.74Nmï¿½ï¿½=0.

084mEPE=1/2ï¿½9.74ï¿½0.084ï¿½=0.

034 JForceF=-2.94mGPE=0KE=0Bungee Cord 1: Energies experienced for the Maximum weight/mass.At the start of the jumpGPE=mghm=93.135kgg=9.8msï¿½ï¿½h=32.84mGPE=93.135ï¿½9.8ï¿½32.

84=29973.82 JEPE=0KE=0Last moment of Freefallvï¿½=uï¿½ + 2asu=0msï¿½ï¿½a=9.8msï¿½ï¿½s=17.

12mv==18.32 msï¿½ï¿½KE=1/2mvï¿½m=93.135kgv=18.32 msï¿½ï¿½KE=1/2ï¿½93.135ï¿½18.32ï¿½=15629.096 JGPE=mghm=93.135kgg=9.

8msï¿½ï¿½h=15.72mGPE=93.135ï¿½9.8ï¿½15.

72=14348.01 JEPE=0At maximum extensionEPE=1/2kï¿½k=58.44Nmï¿½ï¿½=15.72mEPE=1/2ï¿½58.44ï¿½15.72ï¿½=7220.79 JForceF=-(k+c)F=58.44-5.

95k=58.44Nmï¿½ï¿½=15.72mF=-(58.44ï¿½15.72-5.95)=-912.72NGPE=0KE=0At minimum extensionEPE=1/2kï¿½k=58.44Nmï¿½ï¿½=3.

36mEPE=1/2ï¿½58.44ï¿½3.36ï¿½=329.882 JForceF=-912.

72mGPE=0KE=0Bungee Cord 2: Energies experienced for the Maximum weight/mass.At the start of the jumpGPE=mghm=155.22gg=9.8msï¿½ï¿½h=32.84mGPE=155.

22ï¿½9.8ï¿½32.84=49954.76 JEPE=0KE=0Last moment of Freefallv =18.32 msï¿½ï¿½KE=1/2mvï¿½m=155.22kgv=18.32 msï¿½ï¿½KE=1/2ï¿½155.22ï¿½18.

32ï¿½=26042.19 JGPE=mghm=155.22kgg=9.8msï¿½ï¿½h=15.72mGPE=155.22ï¿½9.8ï¿½15.72=23912.

57 JEPE=0At maximum extensionEPE=1/2kï¿½k=97.4Nmï¿½ï¿½=15.72mEPE=1/2ï¿½97.4ï¿½15.72ï¿½=12034.

67 JForceF=-(k+c)F= F=97.4 – 9.928k=97.4Nmï¿½ï¿½=15.72mF=-(97.4ï¿½15.

72-9.928)=-1521.2 NGPE=0KE=0At minimum extensionEPE=1/2kï¿½k=97.

4Nmï¿½ï¿½=3.36mEPE=1/2ï¿½97.4ï¿½3.

36ï¿½=549.8 JF=-1521.2NGPE=0KE=0Summary of Analysis and DiscussionAs a whole, expected results were gathered from the experiments, which could then be analysed and used as the basis of calculations. However, some of the results obtained from both experiment one and two were possibly inaccurate and could be improved. Consequently the calculations would have inaccuracies.

Nevertheless the miscalculations on observed results were minor and would not have a great impact on the calculations. Future changes or improvements to this investigation to ensure greater accuracy could be to use more efficient means of measuring and observing data. Measurements of the mass holder could have been more accurate with the use of a vernier calliper. The extra mass of the blu tac and plastic lid container was not taken into account, although it may have had an effect on the data recorded by the motion sensor.

Instead of observing and approximating results from the position-time graph, obtained from the motion sensor, an alternative method could have been used to find the exact positions for different times. Repeating these experiments would help improve consistency and accuracy.Main results:Strands in ParallelOriginal LengthMassConstantEquationExperiment 1-minimum extensions242.8cm50g to 500g9.

74 Nmï¿½ï¿½F= 9.74 +2.123 (effective for masses of 300g or above)Strands in ParallelOriginal LengthMassConstantMaximum ExtensionMinimum ExtensionEquationExperiment 2-maximum extensions242.8cm300g9.74 Nmï¿½ï¿½39.3cm8.4cmF=9.

74 -0.887(effective for masses of 300g or above)Each Bungee cord consists of 17.12m long, strands in parallel.CordStrands in ParallelLengthConstantEquation (maximum extensions)Minimum Weight/MassMaximumWeight/MassMin-Max extensionsFrom exp. 2217.12m9.74 Nmï¿½ï¿½F=9.74 -0.887(effective for masses of 300g or above)152.127Nï¿½15.52kg11217.12m58.44Nmï¿½ï¿½F=58.44-5.95(effective for masses of 1.8kg or above)286.25Nï¿½29.21kg912.72Nï¿½93.135kg5m-15.72m22017.12m97.4Nmï¿½ï¿½F=97.4 -9.928(effective for masses of 3kg or above)912.72Nï¿½93.135kg1521.2Nï¿½155.22kg9.47m-15.72mCord 1 and 2 (max weight):Length of cord: 17.12mMax extension of cord: 15.72mMin extension: 3.36mLength of extended cord: 32.84mThe energies involved with a bungee jump for the maximum weight/mass.CordStart of JumpLast moment of Free fallMaximum extensionMinimum extensionFrom exp.20.3gGPE= 2.414 JEPE=0KE=0v =2.896 msï¿½ï¿½t =0.295sKE =1.258 GPE=1.155 JEPE=0EPE =0.752 JF=-2.94NGPE=0KE=0EPE=0.034 JF=-2.94mGPE=0KE=0193.135kg912.72NGPE=29973.82 JEPE=0KE=0v=18.32 msï¿½ï¿½KE=15629.096 JGPE=14348.01 JEPE=0EPE=7220.79 JF=-912.72NGPE=0KE=0EPE=329.882 JF=-912.72mGPE=0KE=02155.22kg1521.2NGPE=49954.76 JEPE=0KE=0v=18.32 msï¿½ï¿½KE=26042.19 JGPE=23912.57 JEPE=0EPE=12034.67 JF=-1521.2 NGPE=0KE=0EPE=549.8 JF=-1521.2NGPE=0KE=0Conclusion:This experiment was conducted in order to investigate the important features involved with bungee jump, with the intention of using the results and calculations as a foundation to produce suitable bungee cords, accommodating a range of clients, jumping from the Story Bridge. The association of Hooke’s law and the experimented elastic cord was investigated by trialling different masses and calculating extensions. From the collected results, a model bungee jump was created and a position-time graph was accumulated. These results were carefully analysed and utilised as the basis of further calculations and predictions. Equations were formed to determine the parameters of actual bungee cords which could be applied to the Storey Bridge. The maximum motions and energies experienced were calculated. This experiment could be altered to make further improvements and to increase accuracy of the gathered results. Predominantly, the expected results were derived from this experiment, and the aim was fulfilled.