(a) 2 Hz
The frequency of the nth-harmonic is given by
[tex]f_n = n f_1[/tex]
where
[tex]f_1[/tex] is the fundamental frequency
Therefore, the frequency of the third harmonic of the A ([tex]f_1 = 440 Hz[/tex]) is
[tex]f_3 = 3 \cdot f_1 = 3 \cdot 440 Hz =1320 Hz[/tex]
while the frequency of the second harmonic of the E ([tex]f_1 = 659 Hz[/tex]) is
[tex]f_2 = 2 \cdot f_1 = 2 \cdot 659 Hz =1318 Hz[/tex]
So the frequency difference is
[tex]\Delta f = 1320 Hz - 1318 Hz = 2 Hz[/tex]
(b) 2 Hz
The beat frequency between two harmonics of different frequencies f, f' is given by
[tex]f_B = |f'-f|[/tex]
In this case, when the strings are properly tuned, we have
- Frequency of the 3rd harmonic of A-note: 1320 Hz
- Frequency of the 2nd harmonic of E-note: 1318 Hz
So, the beat frequency should be (if the strings are properly tuned)
[tex]f_B = |1320 Hz - 1318 Hz|=2 Hz[/tex]
(c) 1324 Hz
The fundamental frequency on a string is proportional to the square root of the tension in the string:
[tex]f_1 \propto \sqrt{T}[/tex]
this means that by tightening the string (increasing the tension), will increase the fundamental frequency also*, and therefore will increase also the frequency of the 2nd harmonic.
In this situation, the beat frequency is 4 Hz (four beats per second):
[tex]f_B = 4 Hz[/tex]
And since the beat frequency is equal to the absolute value of the difference between the 3rd harmonic of the A-note and the 2nd harmonic of the E-note,
[tex]f_B = |f_3-f_2|[/tex]
and [tex]f_3 = 1320 Hz[/tex], we have two possible solutions for [tex]f_2[/tex]:
[tex]f_2 = f_3 - f_B = 1320 Hz - 4 Hz = 1316 Hz\\f_2 = f_3 + f_B = 1320 Hz + 4 Hz = 1324 Hz[/tex]
However, we said that increasing the tension will increase also the frequency of the harmonics (*), therefore the correct frequency in this case will be
1324 Hz
