INTERNATIONAL COLLEGEIB Diploma program

Physics Lab Report

Standing Waves Experiment

Candidate Number:

Omar Zeinelabidine

May 2015

Standing wavesRelevant IB Topic: 4.6 HL and 9.2 SL

In this IB lab you will be assessed using the following criteria:

Data collection and Processing; Conclusion and Evaluation.

Aim:In this experiment you have to find the relationship that exists between the wave speed and the tension of a spring and to determine the mass per unit length of the string in kg.m-1.

Apparatus:Signal generator, vibrator, string, slotted masses, meter ruler, pulley, and a ramp.

Diagram:Signal generatorString PulleyVibratorRampWeight

Method and data collection:Adjust the frequency of the vibrator to 100 Hz.Put a 40 g mass on the string and adjust the length of the string until you get a set of standing waves (as shown in the diagram). Record the frequency and the wavelength of the resonant standing wave.Repeat the above steps for different masses between 40g and 200g (at least 6 trials). Theory:The speed of the wave in a string is given by the relation where is the tension in the string andis the mass per unit length. Data processing and presentationPresent the data you collect in an appropriate form.Process the data collected to help you: a- Verify the relationship between the tension in the string and the speed of the wave.b- Calculate the mass per unit length of the string in use through graphical manipulation.Write down all necessary details to reach a final answer.

Aim: To calculate a strings mass per unit length ( in Kg.m-1 by using the strings velocity and tension.

Materials: Set of masses (10g, 20g, 40g, 50g, 100g) 2 meter ruler A vibrator A signal generator Pulley Desk (Ramp)

Variables:

Procedure:

1. Adjust the frequency of the vibrator to 100 Hz.2. Record the mass and the length of the string provided on a little piece of paper with the string. This will be used to compare results from calculation and graphing with the actual mass per unit length.3. Place a 100 g mass on the end of the string, should be under the pulley and not in contact with the ground.4. Adjust the length of the string until 4 spindle like patterns appear.5. Record the length of the string using a 2m ruler (take the beginning of the pulley as a reference for when the string ends).6. Calculate the tension of the string by using Newtons first law (since mass is at rest) =0. Hence 0=mg + T. Which implies T=mg7. Calculate the velocity of the wave by using the equation v= f, where f is the frequency and is the wavelength. Each spindle like pattern is half a wavelength hence there are two wavelengths. To find one wavelength divide the length of the string by two.8. Using the equation calculate the mass per unit length.9. Take three trials for each mass by repeating the same steps above.10. Repeat the same steps above using masses in between 100 and 200 grams. Take at least 6 masses for more accurate results.Figure One: A clear spindle like pattern.

Figure Two: Two spindle like patterns.

Data Collection:

For each mass a length was measured, three trials were taken for each mass. An overall of 9 masses were taken and placed in the table below.

Table One: Collected Data for Mass and LengthMass (m)/g gLength of string (l) in cm 0.5 cm

Trial OneTrial TwoTrial Three

100100.0101.2100.8

120106.5107.8107.2

130112.3110.9111.6

140118.0117.7117.2

150120.9121.2120.7

160125.0125.2125.5

170129.1129.2129.4

180133.2133.7133.1

190134.5133.2134.3

Remark: Since the masses used are fixed, then the uncertainty on the mass is zero.

Data Processing:

The average of the three trials of the lengths must be taken for each mass. This is can be calculated through the equation . the uncertainty can be calculating the difference between the average the furthest value from it. SI units must also be taken hence the grams should be converted to kg by dividing the value by 103, and the length should be divided by 102 to achieve a value in meters (m). After calculating place the values in a table.

Table Two: Masses and average lengths with uncertaintyMass placed at end of string (m)/kg .00 kgAverage length of string (l)/cm Average length of string (l)/m

0.10100.7 0.71.007

0.12107.2 1.072

0.13111.6 1.116

0.14117.6 1.176

0.15120.9 1.209

0.16125.2 1.252

0.17129.2 1.292

0.18133.3 1.333

0.19134.0 1.340

Tension= mg (using Newtons first law) let g be 9.81 for more accurate results.

V= f hence we need lambda which is the length of two spindle like patterns. 2 wavelengths is equal to the length of string so wavelength is half the length of string.To calculate the uncertainty of the velocity we use the following formula:

The frequency has no uncertainty due to the equipment used.

So hence

Sample for the first values at mass= 0.1kg: ) = 0.4 m.s-1

As shown, the value of the wavelength will cancel in the numerator and denominator and we are left with 100 multiplied by the uncertainty of the wavelength. Place the values in a table

Table Three: Values For Tension, Wavelength, and Velocity Mass (m)/kg kgTension of string (T)/kg.m.s-2 kg.m.s-2Wavelength ()/m with uncertainty

Velocity of wave (v)/m.s-1 with uncertainty

0.100.980.504 50.4

0.121.180.536 53.6

0.131.270.558 55.8

0.141.370.589 58.9

0.151.470.603 60.3

0.161.570.626 62.6

0.171.670.646 64.6

0.181.770.667 66.7

0.191.860.670 67.0

Since the equations states that then we can square both sides and the formula will become v2 = . We can plot V vs. T and expect a non linear function to confirm the formula listed above.

As shown there is a non-linear line which means the equation is correct.The Velocity must be squared hence the velocities are squared and placed in a table.

To calculate the uncertainty of the velocity squared, using the first trial as an example:

Let K= v2 Then

Example: )= 40.3.

Uncertainties should be used in one significant digit hence the value above should become 40.

Place the values in a table

Table Four:Table of Tensions with the squares of the velocitiesTension of string kg.m.s-2Velocity of wave (v) m.s-1Velocity squared (v2) in m.s-1 with uncertainty

ValueUncertainty

0.9850.4254040

1.1853.6287340

1.2755.8311440

1.3758.9346920

1.4760.3363620

1.5762.6391930

1.6764.6417310

1.7766.7444930

1.8667.0448950

Using the formula given above: v2= we can rearrange and isolate the mass per unit length and the obtained equation would be Since we have 9 different Tensions and velocities then there shall be 9 mass per unit lengths, the values were placed in a table.

Table Five: Results for mass per unit length 13.86 x 10-4 kg.m-1

24.10 x 10-4 kg.m-1

34.10 x 10-4 kg.m-1

43.96 x 10-4 kg.m-1

54.05 x 10-4 kg.m-1

64.01 x 10-4 kg.m-1

74.00 x 10-4 kg.m-1

83.98 x 10-4 kg.m-1

94.14 x 10-4 kg.m-1

To find the final mass per unit length we must find the average of the 9 values by adding them and dividing the value by 9.

(3.86+4.10+4.10+3.96+4.05+4.01+4.00+3.98+4.14)(10-4) = 4.02 x 10-4

The uncertainty is the value furthest away from the average then subtracted by the average. Hence the value is 0.14 x10-4

The final value for the mass per unit length of the string is: (4.0 ) 10-4 kg.m-1

Graphing:

Using the formula v2 = we can expand it into =, this since then the formula can be rearranged into 4mg= , since we isolate the dependent variable then l2= () m for the slope to have the mass per unit length in it.

We must square the length of the string, and find its uncertainty.

To find its uncertainty: let z= (l)2

Then

For example:

Repeat for the other lengths and place the values in a table

Table Six: Table of mass and the length squared

Mass used (m)/kg kgLength of string (l)/m with uncertainty

The length (l)2/m2 with uncertainty

ValueUncertainty

0.101.007 1.010.01

0.121.072 1.150.02

0.131.116 1.250.02

0.141.176 1.390.01

0.151.209 1.450.01

0.161.252 1.570.01

0.171.292 1.670.01

0.181.333 1.780.01

0.191.340 1.800.02

Plot the values in a graph with the length as the dependent on the y-axis and the mass on the x-axis as the independent.

The slope is 9.453 which is equal to

Then 9.453 f2 = 4g

= 4.15 x10-4 kg.m-1

Using the maximum slope and repeating the step above the calculated mass per unit length is 4.07 x 10-4 kg.m-1

Using the minimum slope and repeating the step above the calculated mass per unit length is 4.59 x 10-4 kg.m-1

Uncertainty is the (maximum value minimum value)/2 = 0.5 x 10-4

Hence the final value is= (4.2 )(10-4) kg.m-1

Conclusion:

There were very few mistakes in this experiment due to a very small uncertainty in the length squares. The recorded mass per unit length recorded in the lab was 0.84x10-3kg 2m = 4.2 x 10-4, which was very close to, the achieved values from both the calculation and the graphing points of view. Gladly both results are accurate and precise which shows how well the experiment was carried out without many errors. However errors can always exist and must be listed.

The uncertainty was x100=5% uncertainty to the real value, which displays the accuracy.

Possible errors:

Miscalculating some values Reading from the incorrect places on ramp 4 standing waves were not fully evident Mistake in procedure Rounding in incorrect places Incorrect Gravitational field strength is used

Possible improvements:

Take more values Take more trials Ask teacher to confirm procedure Find out the exact value of the gravitational field strength before starting the experiment