Answer:
The frequency of note E is:
f(E) = 330 Hz
The frequency of note C is:
f(C) = 262 Hz
The ratio of the frequency of note E to the frequency of note C is just the quotient of these two frequencies:
r = f(E)/f(C) = 330Hz/262Hz = 330/262 = 1.26
Now, we want to find how man E waves will fit in the length of four C waves.
Note that here the word "length" is used, so we need to work with the wavelengths, not with the frequencies.
For waves, we have the relationship:
v = f*λ
where:
v = velocity (in this case, velocity of the sound = 343 m/s)
f = frequency
λ = wavelength.
So, the length of a single E wave is:
λ(E) = (343 m/s)/(330 1/s) = 1.04 m
And the length of a single C note is:
λ(C) = (343 m/s)/(262 1/s) = 1.30 m
In four C waves, the length is:
4*λ(C) = 4*1.30m = 5.2m
The number of E waves that fit in the length of four C waves is equal to the quotient between the length of four C waves and one E wave:
N = (4*λ(C))/(λ(E) ) = (5.2 m)/(1.04m) = 5.14
So we can fit 5 E waves into four C waves.
