E303 AJS

Mapúa Institute of Technology Department of Physics

Experiment 303 TRANSVERSE WAVE: FREQUENCY OF VIBRATION

Name: Saccuan, April Jem H. Program/Year: CpE/2 Course Code/Section: PHY12L/A1 Student No.: 2012100116 Group No.: 5 Seat No.: 502 Date of Performance: January 28, 2014 Date of Submission: February 4, 2014

Prof. Bobby Manlapig Instructor

GRADE

GUIDE QUESTIONS 1. What effect does increasing the tension have on the number of segment formation? Explain your answer If all other factors (length of the string, frequency, and linear mass density) to remain constant, increasing the tension would decrease the number of segment formation. Since the number of segments formed is directly proportional to the square root of the frequency, length, and linear mass density and inversely proportional to the square root of the tension, therefore increasing the tension would decrease the number of segment formation. 2. If the length of the string and the tension in it are constant, what effect does changing the frequency have on the number of segments formed? Explain your answer. If the length of the string and the tension in it are constant, changing the frequency would directly change the number of segments formed. Since the number of segments formed is directly proportional to the square root of the frequency, length, and linear mass density and inversely proportional to the square root of the tension, therefore increasing the frequency would increase the number of segments formed and decreasing the frequency would decrease the number of segments formed. 3. All the six strings on a guitar are of the same length but have different frequencies. Give 3 characteristic differences in the strings that give them differences in pitch. Differences in pitch could be due to the string’s linear mass density (thickness), tension, and the velocity of propagation. Problems: 1. A string has a mass per unit length of 3 x 10-3 g/cm and is attached to an electrically driven vibrator of frequency 100 Hz. How long is the string if the number of segments produced is 2 when under a tension of 1.96 N? Given: Solution: 𝑔 −3 𝑛 𝑇 𝜇 = 3 𝑥 10 𝑓= √ 𝑐𝑚 2𝐿 𝜇 𝑓 = 100 𝐻𝑧 𝑛 𝑇 2 196000 𝑛=2 𝐿= √ = √ 2𝑓 𝜇 2(100) 3 𝑥 10−3 𝑇 = 1.96 𝑁 = 196000 𝑑𝑦𝑛𝑒𝑠 𝐿 = 80.83 𝑐𝑚 Required: 𝐿 2. A 2-meter long wire vibrates with a frequency of 330 Hz when the tension is 500 N. What is the new frequency if the tension on the wire is doubled? Given: Solution: When 𝑇2 = 2𝑇1, 𝐿 =2𝑚 𝑛 𝑇 𝑛2 𝑇2 𝑓= √ 𝑓 = ( ) √ 2 2𝐿 𝜇 𝜇 4𝐿2 𝑓1 = 330 𝐻𝑧 𝑛2 𝑇 𝑇1 = 500 𝑁 𝑛2 2𝑇 𝑓 2 = ( 2) 𝑓2 = √ ( 12 ) 𝜇 4𝐿 Required: 𝑓2

𝑛2 𝜇 𝑛2 𝜇

= =

𝜇

4𝑓 2 𝐿2 𝑇 4(330)(2)2 500

4𝐿

𝑓2 = √ 3484.8 ( = 3484.8

𝑓2 = 466.69 𝐻𝑧

2(500) 4(2)2

)

MAP UA INSTITUTE OF TECHNOLOGY DEPARTMENT OF PHYSICS EXPERIMENT 303 : TRANSVERSE WAVE: FREQUENCY OF VIBRATION Name Program/Year Subject/Section

Saccuan, April Jem H. CpE/2 PHY12L/A1

Group No. Seat No. Date.

5 502 01/28/14

DATA and OBSERVATIONS TABLE 1. Determining the frequency of vibration (constant linear mass density)

Diameter of wire = 0.022 in linear mass density of wire, 𝜇 = 0.0184 g/cm (Please refer to Table 2 for the different size of the string and its equivalent linear mass density)

TRIAL

tension, T (mass added + mass of pan) x 980 cm/s2

number of segments, n

length of string with complete number of segments, L

frequency of vibration, 𝑓=

𝑛 𝑇 √ 2𝐿 𝜇

1

53900 dynes

5

45 cm

95.09 Hz

2

63700 dynes

5

45 cm

103.37 Hz

3

73500 dynes

5

45 cm

111.04 Hz

4

83300 dynes

5

45 cm

118.21 Hz

5

93100 dynes

4

45 cm

99.97 Hz

average frequency of vibration

105.54 Hz

actual value of frequency of vibration

105 Hz

% error

0.51 %

MAP UA INSTITUTE OF TECHNOLOGY DEPARTMENT OF PHYSICS EXPERIMENT 303 : TRANSVERSE WAVE: FREQUENCY OF VIBRATION Name Program/Year Subject/Section

Saccuan, April Jem H. CpE/2 PHY12L/A1

Group No. Seat No. Date.

5 502 01/28/14

DATA and OBSERVATIONS TABLE 1. Determining the frequency of vibration (variable linear mass density)

number of segments, n

length of string with complete number of segments, L

frequency of vibration,

diameter of wire

linear mass density, 𝜇

tension, T (mass added + mass of pan) x 980 cm/s2

1

0.010 in

0.0039 g/cm

93100 dynes

2

45 cm

108.58 hz

2

0.014 in

0.0078 g/cm

93100 dynes

3

45 cm

115.16 hz

3

0.017 in

0.0112 g/cm

93100 dynes

4

45 cm

128.14 hz

4

0.020 in

0.0150 g/cm

93100 dynes

4

45 cm

110.73 hz

5

0.022 in

0.0184 g/cm

93100 dynes

4

45 cm

99.97 hz

TRIAL

average frequency of vibration

𝑓=

𝑛 𝑇 √ 2𝐿 𝜇

107.50 hz

actual value of frequency of vibration

105 hz

% error

2.38 %

Approved by:

Prof. Bobby Manlapig instructor

01/28/14 date

SAMPLE COMPUTATION Table 1. Determining the frequency of vibration (constant linear mass density) 𝜇 = 0.0184

𝑔 𝑐𝑚

𝑇 = 55𝑔 𝑥 980

𝑐𝑚 = 53900 𝑑𝑦𝑛𝑒𝑠 𝑠2

𝑛=5 𝐿 = 45 𝑐𝑚 𝑓=

𝑛 𝑇 √ 2𝐿 𝜇

𝑓=

5 53900 √ 2(45) 0.0184

= 95.09 Hz

Average frequency of vibration = 105.54 Actual value of frequency of vibration = 105 Hz % error =

| 𝐴.𝑉.−𝐸.𝑉.| 𝑥 𝐴.𝑉.

100% =

| 105 −105.4 | 105

𝑥 100% = 0.51%

Table 2. Determining the frequency of vibration (variable linear mass density)

𝜇 = 0.0039

𝑔 𝑐𝑚

𝑇 = 95𝑔 𝑥 980

𝑐𝑚 = 93100 𝑑𝑦𝑛𝑒𝑠 𝑠2

𝑛=2 𝐿 = 45 𝑐𝑚 𝑓=

𝑛 𝑇 √ 2𝐿 𝜇

𝑓=

2 93100 √ 2(45) 0.0039

= 108.58 Hz

Average frequency of vibration = 107.50 Actual value of frequency of vibration = 105 Hz % error =

| 𝐴.𝑉.−𝐸.𝑉.| 𝑥 𝐴.𝑉.

100% =

| 105 −107.50 | 105

𝑥 100% = 2.38%

ANALYSIS For the first part of the experiment, the frequency of vibration is determined by maintaining trials at constant linear mass density with varying tension using a guitar string. From the data, it is shown that as the tension becomes greater, the number of segments formed decreases and at the same time, the frequency increases. If we would to consider all other factors to remain constant and only the tension to change, increasing the tension would decrease the number of segment formation since the number of segments formed is directly proportional to the square root of the frequency, length, and linear mass density and inversely proportional to the square root of the tension.

For the second part of the experiment we kept constant the tension of the strings and the length while varying the linear mass density by using different guitar strings. Seeing the data in Table 2, it shows that as linear mass density increases, its frequency will decrease when observed under the same number of segments formed. Using the relationship stated in equation 3 of the manual, the data is consistent with the theory about their directly proportional relationship. A 0.51% and 2.38% errors for part 1 and part 2 respectively, it means that errors rise in the performance of the experiment. Some errors would come from not having a precise and accurate measurements in the number of segments, and the length. The string must be ensured to be free to vibrate without obstructions.

CONCLUSION Transverse waves shows the vibration of particles of the medium perpendicular to the direction of wave propagation or motion along the medium. In this experiment we determined the frequency of stretched strings, taking consideration the factors like tension and linear mass density to have an effect to the frequency of the strings producing different pitch. Formulating the frequency formula, it shows the relationship that the tension is directly proportional to the frequency and inversely proportional to the number of segments. On the other hand, the linear mass density is inversely proportional to frequency and directly proportional to the number of segments. After conducting the experiment, the objectives are met and the data gathered are consistent with the theory. Sources of error came from inaccurate measurement of length and number of segments.