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Sagot :
### (4a) Two factors that affect the velocity of sound in air (1 mark)
1. Temperature: Higher temperatures result in a higher velocity of sound. This is because the molecules in the air move faster at higher temperatures, facilitating the propagation of sound waves.
2. Humidity: Increased humidity increases the velocity of sound. Water vapor in the air reduces the density of the air, allowing sound waves to travel faster.
### (4b) Solving the problem:
Given:
- Fundamental frequency of the open pipe ([tex]\( f_{\text{open}} \)[/tex]) = 220 Hz
- Velocity of sound in air ([tex]\( v \)[/tex]) = 330 m/s
#### (i) Length of each of the pipes
For an open pipe, the relationship between the fundamental frequency and the length of the pipe is given by:
[tex]\[ f_{\text{open}} = \frac{v}{2 L_{\text{open}}} \][/tex]
Solving for [tex]\( L_{\text{open}} \)[/tex]:
[tex]\[ L_{\text{open}} = \frac{v}{2 f_{\text{open}}} \][/tex]
Substitute the given values:
[tex]\[ L_{\text{open}} = \frac{330 \, \text{m/s}}{2 \times 220 \, \text{Hz}} = 0.75 \, \text{m} \][/tex]
Next, we find the fundamental frequency of the closed pipe under the condition that its first overtone matches the first overtone of the open pipe.
For a closed pipe, the third harmonic (first overtone) is the same as the second harmonic (first overtone) of the open pipe:
[tex]\[ f_{\text{closed, 3rd}} = 2 \times f_{\text{open}} \][/tex]
The fundamental frequency ([tex]\( f_{\text{closed, 1st}} \)[/tex]) of the closed pipe is:
[tex]\[ f_{\text{closed, 1st}} = \frac{f_{\text{closed, 3rd}}}{3} = \frac{2 \times f_{\text{open}}}{3} \][/tex]
Substitute [tex]\( f_{\text{open}} = 220 \)[/tex] Hz:
[tex]\[ f_{\text{closed, 1st}} = \frac{2 \times 220 \, \text{Hz}}{3} = 146.67 \, \text{Hz} \][/tex]
For a closed pipe, the fundamental frequency is related to the length by:
[tex]\[ f_{\text{closed, 1st}} = \frac{v}{4 L_{\text{closed}}} \][/tex]
Solving for [tex]\( L_{\text{closed}} \)[/tex]:
[tex]\[ L_{\text{closed}} = \frac{v}{4 f_{\text{closed, 1st}}} \][/tex]
Substitute the known values:
[tex]\[ L_{\text{closed}} = \frac{330 \, \text{m/s}}{4 \times 110 \, \text{Hz}} = 0.75 \, \text{m} \][/tex]
Thus, the lengths of the open and closed pipes are:
[tex]\[ L_{\text{open}} = 0.75 \, \text{m} \quad \text{and} \quad L_{\text{closed}} = 0.75 \, \text{m} \][/tex]
#### (ii) Fundamental frequency of the closed pipe
We have already determined that:
[tex]\[ f_{\text{closed, 1st}} = 110 \, \text{Hz} \][/tex]
### Summary:
- Length of the open pipe: [tex]\( L_{\text{open}} = 0.75 \, \text{m} \)[/tex]
- Length of the closed pipe: [tex]\( L_{\text{closed}} = 0.75 \, \text{m} \)[/tex]
- Fundamental frequency of the closed pipe: [tex]\( f_{\text{closed, 1st}} = 110 \, \text{Hz} \)[/tex]
1. Temperature: Higher temperatures result in a higher velocity of sound. This is because the molecules in the air move faster at higher temperatures, facilitating the propagation of sound waves.
2. Humidity: Increased humidity increases the velocity of sound. Water vapor in the air reduces the density of the air, allowing sound waves to travel faster.
### (4b) Solving the problem:
Given:
- Fundamental frequency of the open pipe ([tex]\( f_{\text{open}} \)[/tex]) = 220 Hz
- Velocity of sound in air ([tex]\( v \)[/tex]) = 330 m/s
#### (i) Length of each of the pipes
For an open pipe, the relationship between the fundamental frequency and the length of the pipe is given by:
[tex]\[ f_{\text{open}} = \frac{v}{2 L_{\text{open}}} \][/tex]
Solving for [tex]\( L_{\text{open}} \)[/tex]:
[tex]\[ L_{\text{open}} = \frac{v}{2 f_{\text{open}}} \][/tex]
Substitute the given values:
[tex]\[ L_{\text{open}} = \frac{330 \, \text{m/s}}{2 \times 220 \, \text{Hz}} = 0.75 \, \text{m} \][/tex]
Next, we find the fundamental frequency of the closed pipe under the condition that its first overtone matches the first overtone of the open pipe.
For a closed pipe, the third harmonic (first overtone) is the same as the second harmonic (first overtone) of the open pipe:
[tex]\[ f_{\text{closed, 3rd}} = 2 \times f_{\text{open}} \][/tex]
The fundamental frequency ([tex]\( f_{\text{closed, 1st}} \)[/tex]) of the closed pipe is:
[tex]\[ f_{\text{closed, 1st}} = \frac{f_{\text{closed, 3rd}}}{3} = \frac{2 \times f_{\text{open}}}{3} \][/tex]
Substitute [tex]\( f_{\text{open}} = 220 \)[/tex] Hz:
[tex]\[ f_{\text{closed, 1st}} = \frac{2 \times 220 \, \text{Hz}}{3} = 146.67 \, \text{Hz} \][/tex]
For a closed pipe, the fundamental frequency is related to the length by:
[tex]\[ f_{\text{closed, 1st}} = \frac{v}{4 L_{\text{closed}}} \][/tex]
Solving for [tex]\( L_{\text{closed}} \)[/tex]:
[tex]\[ L_{\text{closed}} = \frac{v}{4 f_{\text{closed, 1st}}} \][/tex]
Substitute the known values:
[tex]\[ L_{\text{closed}} = \frac{330 \, \text{m/s}}{4 \times 110 \, \text{Hz}} = 0.75 \, \text{m} \][/tex]
Thus, the lengths of the open and closed pipes are:
[tex]\[ L_{\text{open}} = 0.75 \, \text{m} \quad \text{and} \quad L_{\text{closed}} = 0.75 \, \text{m} \][/tex]
#### (ii) Fundamental frequency of the closed pipe
We have already determined that:
[tex]\[ f_{\text{closed, 1st}} = 110 \, \text{Hz} \][/tex]
### Summary:
- Length of the open pipe: [tex]\( L_{\text{open}} = 0.75 \, \text{m} \)[/tex]
- Length of the closed pipe: [tex]\( L_{\text{closed}} = 0.75 \, \text{m} \)[/tex]
- Fundamental frequency of the closed pipe: [tex]\( f_{\text{closed, 1st}} = 110 \, \text{Hz} \)[/tex]
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