Unit 6 Lab Questions
Guide Questions What effect does increasing the tension have on the number of segment formation? Explain your answer.
According to a theory, changing the tension will definitely change the number of segments. From equation three, the frequency is directly proportional to the number of segments and the tension and inversely proportional to the mass density and the length of the wire. Manipulating the formula to solve for the number of segments, n, is inversely proportional to the tension, thus, increasing the tension produces less number of segments.
If the length of the string and the tension in it are constant, what effect does changing frequency have on the number of segments formed? Explain your answer.
Manipulating equation three
to solve for the number of segments, n, the frequency is directly proportional to the number of segments. If the value of frequency is increased, the number of segments also increases, if the value of frequency is decreased, the number of segments also decreases.
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.
The strings on the guitar differ on their diameter and linear mass density, which are factors that affect the frequency, thus, giving them different pitches. Another factor that varies their pitches is the tension applied on it.
Problems
A string has a mass per unit length of 3x10-3 g/cm and is attached to an electrically driven vibrator of frequency 100Hz. How long is the string if the number of segments produced is 2 when under a tension of 1.96N? µ = 3x10-3 g/cm = 3x10-4 kg/m n = 2 ƒ = 100 Hz T = 1.96N ƒ= n/2L √(T/µ) L= n/(2 ƒ) √(T/µ)
L= 2/(2 (100)) √(1.96/.0003)
L = 0.4427 m A 2-meter long wire vibrates with a frequency of 330 Hz when the tension is 500N. What is the new frequency if the tension on the wire is doubled?
L = 2 m ƒ = 300 Hz T = 500 N
((ƒ^2)/T)A = ((ƒ^2)/T)B = ((330^2)/500) = ((ƒ^2)/1000) ƒ = 466.69 Hz
According to a theory, changing the tension will definitely change the number of segments. From equation three, the frequency is directly proportional to the number of segments and the tension and inversely proportional to the mass density and the length of the wire. Manipulating the formula to solve for the number of segments, n, is inversely proportional to the tension, thus, increasing the tension produces less number of segments.
If the length of the string and the tension in it are constant, what effect does changing frequency have on the number of segments formed? Explain your answer.
Manipulating equation three
to solve for the number of segments, n, the frequency is directly proportional to the number of segments. If the value of frequency is increased, the number of segments also increases, if the value of frequency is decreased, the number of segments also decreases.
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.
The strings on the guitar differ on their diameter and linear mass density, which are factors that affect the frequency, thus, giving them different pitches. Another factor that varies their pitches is the tension applied on it.
Problems
A string has a mass per unit length of 3x10-3 g/cm and is attached to an electrically driven vibrator of frequency 100Hz. How long is the string if the number of segments produced is 2 when under a tension of 1.96N? µ = 3x10-3 g/cm = 3x10-4 kg/m n = 2 ƒ = 100 Hz T = 1.96N ƒ= n/2L √(T/µ) L= n/(2 ƒ) √(T/µ)
L= 2/(2 (100)) √(1.96/.0003)
L = 0.4427 m A 2-meter long wire vibrates with a frequency of 330 Hz when the tension is 500N. What is the new frequency if the tension on the wire is doubled?
L = 2 m ƒ = 300 Hz T = 500 N
((ƒ^2)/T)A = ((ƒ^2)/T)B = ((330^2)/500) = ((ƒ^2)/1000) ƒ = 466.69 Hz