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Experiment to Prove Hooke's Law

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Hooke’s Law

Aim: -To prove Hooke’s law i.e. the extension of the force is directly proportional to the force applied.

- To find the spring constant of the spring.

Apparatus:

  • Clamp Stand
  • Helical Spring
  • Mass Hanger
  • Pointer
  • Meter Ruler
  • Measuring Balance

Method:

-Hang a helical spring from a clamp stand.

-Attach a mass directly to the bottom of the helical spring and record the position of the bottom of the mass hanger relative to a meter ruler.

-Add masses to the spring and record the position of the bottom of the mass hanger.

Safety Precautions:

- Wear safety goggles to prevent any accidents that could occur due to the weights bouncing off the spring.

- Keep a distance from the apparatus.

- Be sure that the spring is tightly attached to the clamp.

- Do not play around with the masses or springs.

Data Collection and Processing

Uncertainty in a measuring balance = ±0.1g

To covert to kg = 0.1÷1000 = ±0.0001kg

Uncertainty in a meter ruler = ±0.05cm

To convert to meters = 0.05 ÷ 100 = ±0.0005m

Formulas

Absolute Uncertainty= Limit of reading÷2

Relative Uncertainty= Absolute Uncertainty ÷ Measured Value

% Uncertainty = Absolute Uncertainty ÷ Measured Value × 100

Force (Newton’s) = Mass (Kg) × Acceleration (ms-²)

Average Extension (cm) = Extension while loading (m) + Extension while unloading (m) ÷ 2

Spring Constant, k (Nm-¹) = Force (Newton’s) ÷ Extension (m)

Elastic Potential Energy (Joules) = 0.5 × Spring Constant × Extension²

Range Of Extension = Extension while loading – Extension while unloading

Random Error = Range of extension ÷ 2

Table 1- Raw Data Table:

Trial No.

Mass

(grams)

±0.1

Mass

(kilograms)

±0.0001

Force Applied

(Newton’s)

F=M×g

±0.0001

Extension While Loading(meters)

±0.0005

Extension While Unloading(meters)

±0.0005

Average Extension =E1+E2÷2 (meters) ±0.001

1

10.2±0.1

0.0102±0.0001

0.100062±0.0001

0.036±0.0005

0.037±0.0005

0.0365±0.001

2

20.4±0.1

0.0204±0.0001

0.200124±0.0001

0.040±0.0005

0.039±0.0005

0.0395±0.001

3

30.6±0.1

0.0306±0.0001

0.300186±0.0001

0.043±0.0005

0.042±0.0005

0.0425±0.001

4

40.8±0.1

0.0408±0.0001

0.400248±0.0001

0.048±0.0005

0.046±0.0005

0.0470±0.001

5

51.0±0.1

0.0510±0.0001

0.500310±0.0001

0.051±0.0005

0.050±0.0005

0.0505±0.001

6

61.2±0.1

0.0612±0.0001

0.600372±0.0001

0.056±0.0005

0.057±0.0005

0.0565±0.001

7

71.4±0.1

0.0714±0.0001

0.700434±0.0001

0.061±0.0005

0.060±0.0005

0.0605±0.001

8

81.6±0.1

0.0816±0.0001

0.800496±0.0001

0.067±0.0005

0.067±0.0005

0.0670±0.001

Calculations for trial 1

Force (Newton’s) = Mass (kg) × Acceleration (ms-²)

= 10.2±0.1 (g) × 9.81 (ms-²)

= 100.062±0.1 (g)

Covert the g to kg: 100.062 ÷ 1000

= 0.100062±0.0001 (kg)

Average Extension = Extension while loading (cm) + Extension while unloading (cm) ÷ 2

= 3.6±0.05 (cm) + 3.7±0.05 (cm)

= 3.65±0.1cm

In meters = 3.65±0.1cm ÷ 100 = 0.0365±0.001m

Table 2- The range of extension and the random error of the experiment:

Trial No.

Extension While Loading(meters)

±0.0005

Extension While Unloading(meters)

±0.0005

Average Extension =E1+E2÷2 (meters) ±0.001

Force Applied

(Newton’s)

F=M×g

±0.0001

Range of Extension

(meters)

±0.0005

Random Error

(meters)

±0.0005

1

0.036±0.0005

0.037±0.0005

0.0365±0.001

0.100062±0.0001

0.001±0.0005

0.0005±0.0005

2

0.040±0.0005

0.039±0.0005

0.0395±0.001

0.200124±0.0001

0.001±0.0005

0.0005±0.0005

3

0.043±0.0005

0.042±0.0005

0.0425±0.001

0.300186±0.0001

0.001±0.0005

0.0005±0.0005

4

0.048±0.0005

0.046±0.0005

0.0470±0.001

0.400248±0.0001

0.002±0.0005

0.001±0.0005

5

0.051±0.0005

0.050±0.0005

0.0505±0.001

0.500310±0.0001

0.001±0.0005

0.0005±0.0005

6

0.056±0.0005

0.057±0.0005

0.0565±0.001

0.600372±0.0001

0.001±0.0005

0.0005±0.0005

7

0.061±0.0005

0.060±0.0005

0.0605±0.001

0.700434±0.0001

0.001±0.0005

0.0005±0.0005

8

0.067±0.0005

0.067±0.0005

0.0670±0.001

0.800496±0.0001

0.000±0.0005

0.0000±0.0005

Calculations for trial 1

Force (Newton’s) = Mass (kg) × Acceleration (ms-²)

= 10.2±0.1 (g) × 9.81 (ms-²)

= 100.062±0.1 (g)

Covert the g to kg: 100.062 ÷ 1000

= 0.100062±0.0001 (kg)

Average Extension = Extension while loading (cm) + Extension while unloading (cm) ÷ 2

= 3.6±0.05 (cm) + 3.7±0.05 (cm)

= 3.65±0.1cm

In meters = 3.65±0.1cm ÷ 100 = 0.0365±0.001m

Range Of Extension = Maximum Value – Minimum Value

= 0.037±0.0005 – 0.036±0.0005

= 0.001±0.005 (m)

Random Error = Range of extension ÷ 2

= 0.001±0.005 ÷ 2

= 0.0005±0.0005 (m)

Table 3- Processed Data Table:

Trial No.

Force Applied

(Newton’s)

F=M×g

±0.0001

Average Extension =E1+E2÷2 (meters) ±0.001

Spring Constant, k (Nm)

% Uncertainty

Elastic Potential Energy (Joules)

% Uncertainty

1

0.100062±0.0001

0.0365±0.001

2.74±2.8%

0.0018251825±8.3%

2

0.200124±0.0001

0.0395±0.001

5.01±2.6%

0.0039084263±7.7%

3

0.300186±0.0001

0.0425±0.001

7.06±2.4%

0.0063760625±7.1%

4

0.400248±0.0001

0.0470±0.001

8.52±2.1%

0.0094103410±6.4%

5

0.500310±0.0001

0.0505±0.001

9.91±2.0%

0.0126364880±6.0%

6

0.600372±0.0001

0.0565±0.001

10.6±1.8%

0.01721974±5.3%

7

0.700434±0.0001

0.0605±0.001

11.6±1.7%

0.02122945±5.0%

8

0.800496±0.0001

0.0670±0.001

11.9±1.5%

0.02670955±4.5%

Calculations for trial 1

Force (Newton’s) = Mass (kg) × Acceleration (ms-²)

= 10.2±0.1 (g) × 9.81 (ms-²)

= 100.062±0.1 (g)

Covert the g to kg: 100.062 ÷ 1000

= 0.100062±0.0001 (kg)

Average Extension = Extension while loading (cm) + Extension while unloading (cm) ÷ 2

= 3.6±0.05 (cm) + 3.7±0.05 (cm)

= 3.65±0.1cm

In meters = 3.65±0.1cm ÷ 100 = 0.0365±0.001m

Spring Constant = Force (Newton’s) ÷ Extension (m)

= 0.100062±0.0001 (N) ÷ 0.0365±0.001 (m)

% Uncertainty for Force = Absolute Uncertainty ÷ Measured Value × 100

= 0.0001 ÷ 0.100062 × 100

= 0.1%

% Uncertainty for Extension = Absolute Uncertainty ÷ Measured Value × 100

= 0.001 ÷ 0.0365 × 100

= 2.7%

Spring Constant = 0.100062±0.1% (N) ÷ 0.0365±2.7% (m)

= 2.74±2.8% Nm-¹

Elastic Potential Energy = 0.5 × Spring Constant × Extension²

= 0.5 × 2.74±2.8% × (0.0365±0.001) ²

= 0.5 × 2.74±2.8% × (0.001332255±5.5%)

= 0.00183±8.3%

Conclusion & Evaluation

Conclusion:

In this experiment, I have been quite successful by proving the aim of the experiment which is Hooke’s Law. The results obtained are slightly incorrect due to any errors as part of the experiment. My calculations were all shown for trial one which whereas follows. In relation to the graph, the line does not pass through the origin as there were uncertainties. The line therefore starts a few cm from the origin on the y axis. The slope in the graph indicates the spring constant. It can be seen that the spring constant value in the graph does not match my result for trial no.1 as I have taken the spring constant value in N/cm. If I take the values in N/m and average all the values of the spring constant from my calculations I will end with a result equal to the gradient or slope of the graph that is 0.227. The units taken for every other value is standard and therefore is correct. My results are reliable as they do result in the Force being proportional to the Extension. I feel that my data is reliable and the graph does show that the extension of the spring directly proportional to the force that is applied to it. We also found that the spring constant and the elastic potential energy increases due to the extension of the spring being proportional to the force.

Evaluation:

I have found that the experiment did have many errors which could have been improved. There were both systematic and random errors involved in the experiment. The meter ruler (uncertainty of ±0.05cm) and the digital balance (uncertainty of ±0.1g) had uncertainty’s which could have altered the accuracy of the results. The experiment also had a parallax error due to the carelessness of me not observing the pointer and the length in the straight path. My equipment was not very accurate as I was given a meter ruler and not an attached ruler. This could have made it very inaccurate as the ruler was leaning over a wall. I could only take one reading per mass, as time management was an issue, which is not reliable as taking more than two readings and averaging the answer will give a more accurate result. The next time I perform this experiment, I will need to make sure that I have at least three readings per mass and should take the average of the three readings to minimize the errors. I should also make sure that the meter ruler is not leaning on a wall and that it is held on by a clamp or that I have the ruler stuck behind the clamp stand. While repeating the experiment one should also put a pointer on the hook to avoid parallax error and get the measurements even more accurate.

  • Wasif Haque

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