Question

A chord consists of notes that sound good together. The C major chord starting at middle C has the following frequencies: C - 262 Hz E - 330 Hz G - 392 Hz determine the ratio of the frequency of E to C. Express the answer in a simple integer ratio. How many E waves will fit in the length of four C waves. a. 2 b. 3

Answer 1

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.