funky problem....

[siddhans]

A certain musical scale has 13 notes, each having a different frequency, measured in cycles per second. In the scale, the notes are ordered by increasing frequency, and the highest frequency is twice the lowest. For each of the 12 lower frequencies, the ratio of a frequency to the next higher frequency is a fixed constant. If the lowest frequency is 440 cycles per second, then the frequency of the 7th note in the scale is how many cycles per second?

a. 440 * root 2
b. 440 * root 2^7
c. 440 * root 2^12
d. 440 * 12root 2^7
e. 440 * 7root2^12



The question stem says that : the ratio of frequency to next higer frequency is fixed constant ...

Doesnt that mean F1/F2 = k

F2/F3 = k^2 and something like that .....???and not F2/F1 = k which is F2 = kF1???

[rijul007]

Option A

[neelgandham]

Let the frequency of the first node be a and
let the fixed constant be x.

So the frequencies of the notes are a, a*x, a*(x^2),a*(x^3),a*(x^4),a*(x^5),a*(x^6) ..............a*(x^12)

So, the question can be rephrased as What is the value of a*(x^6)?

a*(x^12) = 2a [The highest frequency is twice the lowest] => x^12 = 2 => x^6 = Square root(2) - (1)
a = 440 cycles/second (lowest frequency is 440 cycles per second) - (2)

a*(x^6) = 440*Square root(2) - from (1) and (2) above.

IMO Answer : A

[rijul007]

let the frequencies be =
a, b, c, d, ....., m

a = 440
m = 880

a/b = b/c = c/d = ... = l/m

a = b^2/c
c = b^2/a

b/c = c/d
d = c^2/b = (b^2/a)^2/b = b^3/a^2

If you similarly check further terms, you''ll find that the terms are of the form

b^(n-1)/a^(n-2)

so,
13th term(m) = b^12/a^11
880 = b^12/ 440^11
b^12 = 2*440^12
b = 440*2^(1/12)

7th term (g) = b^6/a^5 = (440^6 * 2^(6/12))/440^5
g = 440 * 2^(1/2) = 440 * root 2

Option A[spoiler][/spoiler]