Answer :
Let's break down the problem step by step:
### Given Data:
- Frequency of the third harmonic ([tex]\( f_3 \)[/tex]): 960 Hz
- Length of the string ([tex]\( L \)[/tex]): 0.75 meters
- Speed of sound in air ([tex]\( v \)[/tex]): 343 m/s (standard value)
### a. Wavelength of the third harmonic on the string
For a string fixed at both ends, the harmonics are formed by standing waves. For the third harmonic, there are three half-wavelengths along the length of the string. This can be drawn as follows:
```
|-----|-----|-----|
Note: Each '|' represents a node, whilst each '-----' represents half-wavelengths.
So, for the third harmonic:
[tex]\[ \text{Length of the string} = \frac{3\lambda_3}{2} \][/tex]
Rearrange to find the wavelength ([tex]\(\lambda_3\)[/tex]) of the third harmonic:
[tex]\[ \lambda_3 = \frac{2L}{3} \][/tex]
Substitute the given length of the string:
[tex]\[ \lambda_3 = \frac{2 \times 0.75 \, \text{m}}{3} \][/tex]
[tex]\[ \lambda_3 = \frac{1.5 \, \text{m}}{3} \][/tex]
[tex]\[ \lambda_3 = 0.5 \, \text{m} \][/tex]
Thus, the wavelength of the third harmonic on the string is 0.5 meters.
### b. Wavelength of the third harmonic in air
The wavelength in air ([tex]\(\lambda_{\text{air}}\)[/tex]) can be found using the speed of sound and the frequency:
[tex]\[ \lambda_{\text{air}} = \frac{v}{f}\][/tex]
Substitute the given speed of sound and frequency of the third harmonic:
[tex]\[ \lambda_{\text{air}} = \frac{343 \, \text{m/s}}{960 \, \text{Hz}} \][/tex]
[tex]\[ \lambda_{\text{air}} = 0.357 \, \text{m} \][/tex]
So, the wavelength of the third harmonic in air is approximately 0.357 meters.
### c. Frequency of the first harmonic
The frequency of the harmonics on a string fixed at both ends is given by:
[tex]\[ f_n = n \cdot f_1 \][/tex]
where [tex]\( f_n \)[/tex] is the frequency of the nth harmonic. For the third harmonic ([tex]\( f_3 \)[/tex]):
[tex]\[ f_3 = 3 \cdot f_1 \][/tex]
Rearrange to solve for the fundamental frequency [tex]\( f_1 \)[/tex] (first harmonic):
[tex]\[ f_1 = \frac{f_3}{3} \][/tex]
Substitute the given frequency of the third harmonic:
[tex]\[ f_1 = \frac{960 \, \text{Hz}}{3} \][/tex]
[tex]\[ f_1 = 320 \, \text{Hz} \][/tex]
Thus, the frequency of the first harmonic is 320 Hz.
### d. Listener notice if a speaker plays a tone of 962 Hz while only the third harmonic is playing
When two sound waves of slightly different frequencies are heard together, they create a phenomenon known as beats. The beat frequency ([tex]\( f_{\text{beat}} \)[/tex]) is given by the absolute difference between the two frequencies:
[tex]\[ f_{\text{beat}} = |f_1 - f_2| \][/tex]
Substitute the given frequencies:
[tex]\[ f_{\text{beat}} = |962 \, \text{Hz} - 960 \, \text{Hz}| \][/tex]
[tex]\[ f_{\text{beat}} = 2 \, \text{Hz} \][/tex]
Thus, the listener would notice a beat frequency of 2 Hz, which means they would hear a periodic variation in sound intensity at a rate of 2 beats per second.
### Given Data:
- Frequency of the third harmonic ([tex]\( f_3 \)[/tex]): 960 Hz
- Length of the string ([tex]\( L \)[/tex]): 0.75 meters
- Speed of sound in air ([tex]\( v \)[/tex]): 343 m/s (standard value)
### a. Wavelength of the third harmonic on the string
For a string fixed at both ends, the harmonics are formed by standing waves. For the third harmonic, there are three half-wavelengths along the length of the string. This can be drawn as follows:
```
|-----|-----|-----|
Note: Each '|' represents a node, whilst each '-----' represents half-wavelengths.
So, for the third harmonic:
[tex]\[ \text{Length of the string} = \frac{3\lambda_3}{2} \][/tex]
Rearrange to find the wavelength ([tex]\(\lambda_3\)[/tex]) of the third harmonic:
[tex]\[ \lambda_3 = \frac{2L}{3} \][/tex]
Substitute the given length of the string:
[tex]\[ \lambda_3 = \frac{2 \times 0.75 \, \text{m}}{3} \][/tex]
[tex]\[ \lambda_3 = \frac{1.5 \, \text{m}}{3} \][/tex]
[tex]\[ \lambda_3 = 0.5 \, \text{m} \][/tex]
Thus, the wavelength of the third harmonic on the string is 0.5 meters.
### b. Wavelength of the third harmonic in air
The wavelength in air ([tex]\(\lambda_{\text{air}}\)[/tex]) can be found using the speed of sound and the frequency:
[tex]\[ \lambda_{\text{air}} = \frac{v}{f}\][/tex]
Substitute the given speed of sound and frequency of the third harmonic:
[tex]\[ \lambda_{\text{air}} = \frac{343 \, \text{m/s}}{960 \, \text{Hz}} \][/tex]
[tex]\[ \lambda_{\text{air}} = 0.357 \, \text{m} \][/tex]
So, the wavelength of the third harmonic in air is approximately 0.357 meters.
### c. Frequency of the first harmonic
The frequency of the harmonics on a string fixed at both ends is given by:
[tex]\[ f_n = n \cdot f_1 \][/tex]
where [tex]\( f_n \)[/tex] is the frequency of the nth harmonic. For the third harmonic ([tex]\( f_3 \)[/tex]):
[tex]\[ f_3 = 3 \cdot f_1 \][/tex]
Rearrange to solve for the fundamental frequency [tex]\( f_1 \)[/tex] (first harmonic):
[tex]\[ f_1 = \frac{f_3}{3} \][/tex]
Substitute the given frequency of the third harmonic:
[tex]\[ f_1 = \frac{960 \, \text{Hz}}{3} \][/tex]
[tex]\[ f_1 = 320 \, \text{Hz} \][/tex]
Thus, the frequency of the first harmonic is 320 Hz.
### d. Listener notice if a speaker plays a tone of 962 Hz while only the third harmonic is playing
When two sound waves of slightly different frequencies are heard together, they create a phenomenon known as beats. The beat frequency ([tex]\( f_{\text{beat}} \)[/tex]) is given by the absolute difference between the two frequencies:
[tex]\[ f_{\text{beat}} = |f_1 - f_2| \][/tex]
Substitute the given frequencies:
[tex]\[ f_{\text{beat}} = |962 \, \text{Hz} - 960 \, \text{Hz}| \][/tex]
[tex]\[ f_{\text{beat}} = 2 \, \text{Hz} \][/tex]
Thus, the listener would notice a beat frequency of 2 Hz, which means they would hear a periodic variation in sound intensity at a rate of 2 beats per second.
