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musical sacle has 13 notes each hainvg a different

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Director
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musical sacle has 13 notes each hainvg a different [#permalink]  12 Nov 2005, 10:37
musical sacle has 13 notes each hainvg a different frequency. In scale notes are ordered by increasing frequency and highest note is twice the lowest. For each 12 lower frequencies the ratio of freuency to next higher is fixed constant. If lower frequency = 440 then frequency of 7th note in scale is?

a) 440 Sqr(2)
b) 440 Sqr(2^7)
c) 440 Sqr(2^12)
d)440 Sqr(2^14)
e)440 Sqr(2^16)
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[#permalink]  12 Nov 2005, 16:38
The frequencies are f, f*r, f*r^2, f*r^3,...,f*r^12
Given, f=440 and f*r^12 = 2*f -> r^12 = 2 ---(1)
Asked, f*r^6
Since the answers are in terms of sqrt(2), taking square root on both sides,
r^6 = sqrt(2)
So, f*r^6 = 440*sqrt(2) [choice A]

Last edited by gsr on 13 Nov 2005, 12:02, edited 1 time in total.
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[#permalink]  13 Nov 2005, 11:16
gsr, how did u determine the frequencies? could u please explain your logic? I'm having a hard time grasping the info in the question.
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Re: frequency of 7th note -- [#permalink]  13 Nov 2005, 11:39
cool_jonny009 wrote:
musical sacle has 13 notes each hainvg a different frequency. In scale notes are ordered by increasing frequency and highest note is twice the lowest. For each 12 lower frequencies the ratio of freuency to next higher is fixed constant. If lower frequency = 440 then frequency of 7th note in scale is?

a) 440 Sqr(2)
b) 440 Sqr(2^7)
c) 440 Sqr(2^12)
d)440 Sqr(2^14)
e)440 Sqr(2^16)

Let r be the constant...

1st frequency: 440
2nd: 440*R
3rd: 440*R^2
4th: 440*R^3
5th: 440*R^4
6th: 440*R^5
7th: 440*R^6
..
..
13th: 440*R^12 , but we are also given this frequency is 880

So solving for R we get 440*R^12 = 880, R^12 = 2

R^12 = R^6*R^6 = 2

R^6 = sqrt 2

We are asked 440*R^6, so the answer is 440*sqrt 2

Answer is A
Director
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[#permalink]  13 Nov 2005, 12:06
rsanchez wrote:
gsr, how did u determine the frequencies? could u please explain your logic? I'm having a hard time grasping the info in the question.

The frequencies are in Geometric progression (each successive one is over the other by a constant ratio 'r') -
If f is the 1st freq, then 2nd freq is f*r
Third freq is (f*r)*r = f*r^2
Fourh freq is (f*r^2)*r = f*r^3
13th = f*r^12

We are given two things
1. First frequency = f = 440
2. 13th freq = 2*(1st freq) => f = 2*f^12

Asked is 7th freq = f*r^6

Since the answers are in terms of sqrt(2), taking square root on both sides,
r^6 = sqrt(2)
So, f*r^6 = 440*sqrt(2) [choice A]

Hope this is clear
VP
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[#permalink]  14 Nov 2005, 08:57
how do we know that its geometric and not arithmetic progression ?
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[#permalink]  14 Nov 2005, 09:09
It is a GP since the problem statement states :
Quote:
For each 12 lower frequencies the ratio of freuency to next higher is fixed constant.
.

In a GP, the ration of successive terms is constant. In an AP, the difference is constant.
VP
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[#permalink]  14 Nov 2005, 09:16
justchill wrote:
It is a GP since the problem statement states :
Quote:
For each 12 lower frequencies the ratio of freuency to next higher is fixed constant.
.

In a GP, the ration of successive terms is constant. In an AP, the difference is constant.

_________________

If your mind can conceive it and your heart can believe it, have faith that you can achieve it.

Director
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[#permalink]  14 Nov 2005, 11:59
good job guys... OA is A

[#permalink] 14 Nov 2005, 11:59
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