Let's break down the problem step by step:
### Given Data:
- Frequency of the third harmonic (\(\( f_3 \)\)): 960 Hz
- Length of the string (\(\( L \)\)): 0.75 meters
- Speed of sound in air (\(\( v \)\)): 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:
\(\[ \text{Length of the string} = \frac{3\lambda_3}{2} \]\)
Rearrange to find the wavelength (\(\(\lambda_3\)\)) of the third harmonic:
\(\[ \lambda_3 = \frac{2L}{3} \]\)
Substitute the given length of the string:
\(\[ \lambda_3 = \frac{2 \times 0.75 \, \text{m}}{3} \]\)
\(\[ \lambda_3 = \frac{1.5 \, \text{m}}{3} \]\)
\(\[ \lambda_3 = 0.5 \, \text{m} \]\)
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 (\(\(\lambda_{\text{air}}\)\)) can be found using the speed of sound and the frequency:
\(\[ \lambda_{\text{air}} = \frac{v}{f}\]\)
Substitute the given speed of sound and frequency of the third harmonic:
\(\[ \lambda_{\text{air}} = \frac{343 \, \text{m/s}}{960 \, \text{Hz}} \]\)
\(\[ \lambda_{\text{air}} = 0.357 \, \text{m} \]\)
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:
\(\[ f_n = n \cdot f_1 \]\)
where \(\( f_n \)\) is the frequency of the nth harmonic. For the third harmonic (\(\( f_3 \)\)):
\(\[ f_3 = 3 \cdot f_1 \]\)
Rearrange to solve for the fundamental frequency \(\( f_1 \)\) (first harmonic):
\(\[ f_1 = \frac{f_3}{3} \]\)
Substitute the given frequency of the third harmonic:
\(\[ f_1 = \frac{960 \, \text{Hz}}{3} \]\)
\(\[ f_1 = 320 \, \text{Hz} \]\)
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 (\(\( f_{\text{beat}} \)\)) is given by the absolute difference between the two frequencies:
\(\[ f_{\text{beat}} = |f_1 - f_2| \]\)
Substitute the given frequencies:
\(\[ f_{\text{beat}} = |962 \, \text{Hz} - 960 \, \text{Hz}| \]\)
\(\[ f_{\text{beat}} = 2 \, \text{Hz} \]\)
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.
