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Re: A certain musical scale has has 13 notes, each having a different freq
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Updated on: 13 Jul 2013, 00:28
Pushpinder wrote: Lowest frequency = 440 Highest frequency = 880 Lowest frequency (n) ^12 = Highest frequency N^12 = 2  1 7th note = Lowest Frequency x (n)^6 7th note = 440 x (2)^6/12 Hence the answer is A Pushpinder Ji I couldn't understand from here. Can u tell me please
Originally posted by dasikasuneel on 12 Jul 2013, 22:40.
Last edited by Bunuel on 13 Jul 2013, 00:28, edited 2 times in total.
Edited.



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Re: A certain musical scale has has 13 notes, each having a different freq
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26 Jul 2013, 05:50
Bunuel wrote: dasikasuneel wrote: Pushpinder wrote: A certain musical scale has 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 * sqrt 2 B. 440 * sqrt (2^7) C. 440 * sqrt (2^12) D. 440 * the twelfth root of (2^7) E. 440 * the seventh root of (2^12)
Lowest frequency = 440 Highest frequency = 880 Lowest frequency (n) ^12 = Highest frequency N^12 = 2  1 7th note = Lowest Frequency x (n)^6 7th note = 440 x (2)^6/12 Hence the answer is A Pushpinder Ji I couldn't understand from here. Can u tell me please 1st = \(440\) 2nd = \(440k\) 3rd = \(440k^2\) ... 7th = \(440k^6\) ... 13th = \(440k^{12}=2*440=880\) > \(440k^{12}=880\) > \(k^{12}=2\) > \(k=\sqrt[12]{2}\). Thus, 7th = \(440k^6=440(\sqrt[12]{2})^6=440\sqrt{2}\). Answer: A. Hope it's clear. Hi Bunuel , It says that the ratio of ratio of a frequency to the next higher frequency is a fixed constant. Doesnt that mean f1/f2 = k ?? Just a little lost. Cheers HeirApparent.



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Re: A certain musical scale has has 13 notes, each having a different freq
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26 Jul 2013, 06:40
heirapparent wrote: Bunuel wrote: dasikasuneel wrote: Pushpinder Ji I couldn't understand from here. Can u tell me please 1st = \(440\) 2nd = \(440k\) 3rd = \(440k^2\) ... 7th = \(440k^6\) ... 13th = \(440k^{12}=2*440=880\) > \(440k^{12}=880\) > \(k^{12}=2\) > \(k=\sqrt[12]{2}\). Thus, 7th = \(440k^6=440(\sqrt[12]{2})^6=440\sqrt{2}\). Answer: A. Hope it's clear. Hi Bunuel , It says that the ratio of ratio of a frequency to the next higher frequency is a fixed constant. Doesnt that mean f1/f2 = k ?? Just a little lost. Cheers HeirApparent. Does not matter how you write: f1/f2=constant > f2/f1=1/constant.
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Re: A certain musical scale has has 13 notes, each having a different freq
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24 Sep 2013, 20:12
In a geometric progression, the median is the geometric mean given by SQRT (First * Last). Here, First is 440, Last is 2*440 = 880 and 7th Note is the median, so it's value = SQRT (440*880) = SQRT (440*440*2) = 440*SQRT(2) A is correct.



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Re: A certain musical scale has has 13 notes, each having a different freq
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07 Sep 2017, 08:31
Hello , This is the solution if we take f1/f2=k Given: f1= 440, f 13 =2(440)=880 Also f1/f2 =f2/f3 =f3/f4 =f4/f5 =f5/f6 =f6/f7 =f7/f8 =f8/f9 =f9/f10 =f10/f11 =f11/f12 =f12/f13 = K ( Some constant) need to find f7? we can write f2/f1= 1/k Acc to GP fn= f1(1/k)^n1 Then f7= 440(1/k)^6 and f13= 440(1/k)^12 880=440 (1/k)^12 (1/k)^12= 2 {(1/K)^6}^2= 2 or (1/k)^6= √ ( 2) Substitute the value of (1/k)^6 in equation So f7= 440√ 2 @Banuel is this also a correct approach. Choice A
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Re: A certain musical scale has has 13 notes, each having a different freq
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07 Sep 2017, 10:42
The sequence is in GP a= 440 Ar^n1 =term of GP Now 2ar^11 = ar^131 2= r^121 7th term = ar^6 440*root(2)



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Re: A certain musical scale has has 13 notes, each having a different freq
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11 Dec 2017, 18:17
stoolfi wrote: A certain musical scale has 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 * sqrt 2 B. 440 * sqrt (2^7) C. 440 * sqrt (2^12) D. 440 * the twelfth root of (2^7) E. 440 * the seventh root of (2^12) We are given that a certain musical scale has 13 notes, ordered from least to greatest. We also know that each next higher frequency is equal to the preceding frequency multiplied by some constant. Since the first frequency is 440 cycles per second, the second frequency is 440k, the third is 440k^2, the fourth is 440k^3…the seventh frequency is 440k^6, and the thirteenth frequency is 440k^12. Since the highest frequency is twice the lowest, we can create the following equation: 440 x 2 = 440k^12 2 = k^12 (^12)√2 = k Thus, the seventh frequency is 440((^12)√2)^6 = 440√2. Answer: A
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Re: A certain musical scale has has 13 notes, each having a different freq
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17 Jan 2018, 17:01
stoolfi wrote: A certain musical scale has 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. \(440 * \sqrt 2\)
B. \(440 * \sqrt {2^7}\)
C. \(440 * \sqrt {2^{12}}\)
D. \(440 * \sqrt[12]{2^7}\)
E. \(440 * \sqrt[7]{2^{12}}\) Let k = the multiplier for each successive note. That is, each note is k TIMES the note before it. Then we'll start listing each note: 1st note = 440 cycles per second 2nd note = 440(k) cycles per second 3rd note = 440(k)(k) cycles per second 4th note = 440(k)(k)(k) cycles per second . . . 7th note = 440(k)(k)(k)(k)(k)(k) = 440(k^6) cycles per second . . . 13th note = 440(k)(k)(k)(k)(k)(k)(k)(k)(k)(k)(k)(k) = 440(k^12) cycles per second We're told that the highest frequency is twice the lowest. In other words, 440(k^12) is twice as big as 440We can write: 440(k^12) = (2) 440Divide both sides by 440 to get: k^12 = 2NOTE: Our goal is to find the frequency for the 7th note. In other words, we want to find the value of 440(k^6)Since k^12 = 2, we can rewrite this as (k^6)^2 = 2This means that k^6 = √2 So, the frequency of the 7th note = 440(k^6) = 440√2 Answer = A Cheers, Brent
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A certain musical scale has has 13 notes, each having a different freq
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06 Feb 2019, 21:01
stoolfi wrote: A certain musical scale has 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. \(440 * \sqrt 2\)
B. \(440 * \sqrt {2^7}\)
C. \(440 * \sqrt {2^{12}}\)
D. \(440 * \sqrt[12]{2^7}\)
E. \(440 * \sqrt[7]{2^{12}}\) My reasoning: \(a_1 = 440\) and \(a_13 = 880\) Since we know the frequencies are increasing with a constant ratio we know this is a geometric sequence. Thus \(a_{13} = a_1 * r^{131}\) We can solve for r which is \(2^{\frac{1}{12}}\) With r we can solve for the 7th term \(a_7 = 440 * 2^{\frac{1}{12} * 6}\) \(a_7 = 440 * \sqrt 2\)



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Re: A certain musical scale has has 13 notes, each having a different freq
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30 Mar 2019, 14:27
Bunuel VeritasKarishma isn't the gp for the first 12 numbers of the sequence? how/why are we assuming that 880 is the 13th and the last term of the gp when the question says that the gp is of the 1st 12 scale notes ?




Re: A certain musical scale has has 13 notes, each having a different freq
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