GUIDE QUESTIONS 1. In the length measurement of vibrating string, the segment close to the stylus was not included. Why? The segment close to the stylus was not included since it doesn’t form a complete segment. A complete segment is formed with two anodes. However, the segment close to the stylus has only one anode. It might cause an error in measuring the length of the string when this segment near to the stylus was included. 2. What effect does increasing the tension have on the number of segment formation? Justify your answer. Increasing the tension will make the number of segments to increase also. This is because when there’s an increase in tension, the frequency will increase also. When there’s a higher frequency, more wavelengths are formed. A segment is one-half of a wavelength so when w hen there’s a higher frequency, there will be greater number of segments to be formed. However in the experiment, even though the tension increases, the length of the string must be increased also in order to maintain the frequency thus making the number of segments to be lesser. 3. All strings on the acoustic guitar are of the same length. What characteristic differences do they have that gives them pitch (different frequencies)? Discuss other procedures that can be done to change the frequency. The strings differ in thickness and diameter. There will be a lower frequency when the string has a large diameter. This is because it vibrates slower. The density also affects the frequency. The strings with low pitches are made of more dense material than the strings with high pitches. Tension is applied to the string to stretch and make it tight. Tightening the string gives it a higher frequency while loosening it lowers the frequency.
Problem: 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 vps. How long is the string if the number of segments produced i s 2 when under a tension of 0.2 kg? Given: μ = 3 x 10-3 g/cm, f = 100 vps = 100 Hz, n = 2, T = (200 g)(980 cm/s2) = 196000 g-cm/s2 Required: L = ? Solution:
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 reduced to a factor of 0.5? Given: f 1 = 330 Hz, T1 = 500 N, T2 = 0.5T1 = 0.5(500 N) = 250 N Required: f 2 2 = ? Solution:
√ √
Physics Laboratory Part 3
SAMPLE COMPUTATION
Table 1. Determining the Frequency of Vibration (Constant Linear Mass Density)
Table 2. Determining the Frequency of Vibration (Variable Linear Mass Density)
Trial 1: T = 53900 dynes μ = 0.0039 g/cm L = 35.50 cm n=2
Trial 1: T = 53900 dynes μ = 0.0039 g/cm L = 52.00 cm n=3
= 104.72 Hz
= 107.24 Hz
= 105.77 Hz
= 108.16 Hz
| | | | =
= 0.73%
x 100%
=
x 100%
= 1.67%
Physics Laboratory Part 3
ANALYSIS
Experiment 303 is one of the exciting experiments in Physics laboratory because it involves waves. Here, the properties of transversal waves were observed and examined. When the sine wave generator was turned on, waves were seen moving up and down following a horizontal path. In the first part of the experiment, the frequency of vibration was determined with constant linear mass density and variable tension. It was observed in the gathered data that as the tension increases, the number of segments decreases. This was happened because the length of the string was adjusted and increased to make the frequency to be constant. However, in reality, as the tension increases, the frequency also increases making more segment formations to happen. This means that the number of segments is proportional to the tension. When the tension increased, there will be a higher frequency. The higher the frequency, the more wavelengths formed thus producing greater number of segments. In the second part of the experiment, the frequency of vibration was determined with constant tension and variable linear mass density. It was noticed that an increase in the linear mass density of the string with constant tension will increase the number of segments such that the length of that string was adjusted also to maintain the frequency to be constant. Since there’s greater number of segments, there’s a higher frequency. But during the experiment, it was observed also that when the diameter and the linear mass density of the string increase, the number of segment decreases. The frequency decreases as the linear mass density increases. This means that frequency is inversely proportional to the linear mass density. It was difficult to count the number of segment during the experiment since the waves were not clearly visible. Adjusting the amplitude was very important to see the clear wave motion. There were small percentage errors between the actual and experimental values that were determined. But these values are acceptable since there are minimal amount of errors produced during the experiment.
CONCLUSION
Experiment 303 is guided by the following objectives: (1) to determine the frequency of vibration of a stretched string and (2) to study how the frequency of vibrating string is affected by tension and linear mass density. In this experiment, the properties of transverse waves were observed and studied as well as the factors affecting its frequency. Transverse waves are moving waves in which the particles of the medium undergo oscillations perpendicular to the direction of the energy transfer. Energy is being passed particle to particle which causes them to move up and down. The particles aren’t moving from its place but it is only vibrating up and down. It is only the shape of vibration that is moving to the right. This shape that moves from one particle to the other is what we call the wave motion. In the experiment, the factors affecting the frequency were determined and their relationships were established. Frequency is directly proportional with the tension. When the tension increases, the frequency also increases creating more wavelengths and forms greater number of segments. Frequency is affected also by the linear mass density. It is inversely proportional with the linear mass density of the medium. As the linear mass density increases, the frequency decreases. Frequency is also affected by the length of the medium of the wave. The tension and the density of a string affect the pitch and frequency of the spring. Dense molecules vibrate at slower speeds. The denser the string is, the slower it will vibrate, and the lower its frequency will be. The strings used for low pitches will be made of a more dense material than the strings used for high pitches. Stretching and tightening the string will increase its frequency and pitch while loosening it will decrease its frequency.
Physics Laboratory Part 3
