Ratioed beating freqs

"William Coon"

In your case f2 = 0.6f1

and f1 - f2 = 4400

so 0.4f1 = 4400

giving f1 = 11000

and f2 = 6600

Note:

the true beat frequency is ( f1-f2 )/2 but the ear hears the positive and negative maxima the same.

....... Phil

The frequency difference between the two source sine waves is 4400Hz, so we can write:

df = 4400Hz = |f1 - f2|

where f1 and f2 are the two source frequencies. You can arbitrarily assume that f1 > f2 without loss of generality (since f1 and f2 are interchangeable unknowns) and write:

df = f1 - f2

You also specify that f1 = (1.0/0.6)*f2 = (5/3)*f2 for some choice of f1 and f2. Let r = (1.0/0.6), then you can write:

df = r*f2 - f2 df = f2*(r - 1)

In the example given,

4400Hz = f2*2/3

f2 = 6600Hz and f1 = f2 - 4400Hz = 2200Hz

"Greg Neill"

f1 = f2 +4400 = 11000.

........ Phil

Simple math as in 1+1=2.

Oops. My bad. Thanks.