Question

Piano tuners tune pianos by listening to the beats between the harmonics of two different strings. When properly tuned, the note A should have a frequency of 440 Hz and the note E should be at 659 Hz.
(a) What is the frequency difference between the third harmonic of the A and the second harmonic of the E?

(b) A tuner first tunes the A string very precisely by matching it to a 440 Hz tuning fork. She then strikes the A and E strings simultaneously and listens for beats between the harmonics. What beat frequency between higher harmonics indicates that the E string is properly tuned?

(c) The tuner starts with the tension in the E string a little low, then tightens it. What is the frequency of the E string when she hears four beats per second?

Answer 1

(a) 2 Hz

The frequency of the nth-harmonic is given by

$f_n = n f_1$

where

$f_1$ is the fundamental frequency

Therefore, the frequency of the third harmonic of the A ($f_1 = 440 Hz$) is

$f_3 = 3 \cdot f_1 = 3 \cdot 440 Hz =1320 Hz$

while the frequency of the second harmonic of the E ($f_1 = 659 Hz$) is

$f_2 = 2 \cdot f_1 = 2 \cdot 659 Hz =1318 Hz$

So the frequency difference is

$\Delta f = 1320 Hz - 1318 Hz = 2 Hz$

(b) 2 Hz

The beat frequency between two harmonics of different frequencies f, f' is given by

$f_B = |f'-f|$

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)

$f_B = |1320 Hz - 1318 Hz|=2 Hz$

(c) 1324 Hz

The fundamental frequency on a string is proportional to the square root of the tension in the string:

$f_1 \propto \sqrt{T}$

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):

$f_B = 4 Hz$

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,

$f_B = |f_3-f_2|$

and $f_3 = 1320 Hz$, we have two possible solutions for $f_2$:

$f_2 = f_3 - f_B = 1320 Hz - 4 Hz = 1316 Hz\\f_2 = f_3 + f_B = 1320 Hz + 4 Hz = 1324 Hz$

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