Question

The lowest note on a guitar is typically tuned to an E2. The next lowest string is tuned to an A2 at 110 Hz, which is a fourth (5 half steps) higher than the E2. The next string is tuned to D3, which is a fourth higher than A2.

(a) What are the frequencies of the lowest three open strings on a guitar tuned this way?

(b) If I play the open E2 on the guitar, and then push the A2 string down at the second fret, to increase its frequency by two half steps, the two strings will make a perfect fifth interval (7 half steps). If I do this and then play the two strings, what is the dominant beat frequency I’m most likely to hear?

(c) If I play the open E2 and the open A2, the interval I get is a fourth. If I play these two opens strings together, what is the dominant beat frequency I’m most likely to hear?

Answer 1

(a) Let frequency of E2 be f

Then, frequency of A2 ( fourth higher than E2)  = 4/3 f

Frequency of D3 = 4/3*(4/3*f) =16/9 f

Now, Given 4/3 f = 110 Hz

So, f = 82.5 Hz

therefore,Frequencies (E2, A2 , D3 ) = ( 82.5 Hz, 110 Hz, 146.6 Hz)

(b) It is given that A2 is 5 half steps higher than E2

so , 110 - 82.5 = 5 half steps

27.5 = 5 half steps

1 half step = 5.5 Hz

So, new frequency of A2 = 110 + 2*5.5

= 121 Hz

Beat frequency = 121 - 82.5 = 38.5 Hz

(c) Beat frequency = 110 - 82.5 = 27.5 Hz