a chord consists of notes that sound good together. the c major chord, starting at middle c, has the…

Explanation:

Step1: Find the ratio of frequencies

The frequency of (C) is (f_C = 262\space Hz) and the frequency of (E) is (f_E=330\space Hz). The ratio of frequencies (\frac{f_E}{f_C}=\frac{330}{262}\approx\frac{3}{2}) (after simplification).

We know that the speed of a wave (v = f\lambda) (where (v) is the speed of sound, (f) is the frequency and (\lambda) is the wavelength). Assuming the speed of sound (v) is constant (since the medium is the same), (\lambda=\frac{v}{f}).

Let the wavelength of (C) be (\lambda_C) and of (E) be (\lambda_E). Then (\frac{\lambda_C}{\lambda_E}=\frac{f_E}{f_C}) (because (v = f_C\lambda_C=f_E\lambda_E))

Step2: Calculate the number of (E) waves in 4 (C) waves

Let (n) be the number of (E) waves. We know that (4\lambda_C=n\lambda_E). From (\frac{\lambda_C}{\lambda_E}=\frac{f_E}{f_C}\approx\frac{3}{2}), we can rewrite (4\lambda_C=n\lambda_E) as (n = 4\times\frac{\lambda_C}{\lambda_E})

Substitute (\frac{\lambda_C}{\lambda_E}\approx\frac{3}{2}) into the equation: (n = 4\times\frac{3}{2})

[n=\frac{12}{2}=6] (Wait, no. Let's use another approach.

Since (v = f\lambda), for a given length (L), (L = N\lambda). Also, (v) is constant. So (f_1\lambda_1=f_2\lambda_2). If (L = 4\lambda_C), and (L=n\lambda_E), then (4\lambda_C=n\lambda_E). Since (f_C\lambda_C = f_E\lambda_E), (\lambda_E=\frac{f_C}{f_E}\lambda_C). Substitute into (4\lambda_C=n\lambda_E):

(4\lambda_C=n\times\frac{f_C}{f_E}\lambda_C). Cancel out (\lambda_C) (since (\lambda_C\neq0)). (n = 4\times\frac{f_E}{f_C})

Substitute (f_C = 262), (f_E = 330): (n=4\times\frac{330}{262}\approx4\times1.26\approx 5)

Answer:

d. 5