For standard tuning, concert A is defined to have a frequency of 440 Hz. On a piano. A is five white keys above C, but 9 half steps above C counting both the white and black keys. (See fig. 15.22.) A full octave consists of 12 half steps (semitones). In equally tempered tuning, each half step has the ratio of 1.0595 above the preceding step. (This ratio is the 12th root of 2.0.) a. What is the frequency of A-flat, one half step below A for equal temperament? b. Working down, find the frequency of each succeeding half step until you get down to C. (Carry your computations to four figures, to avoid rounding errors. For each half step, divide by 1.0595.) c. In just tuning, middle C has a frequency of 264 Hz. How does your result in part b compare to this value? d. Working up, find the frequency of C above concert A in equal temperament. Is this frequency twice that obtained in part b for middle C?

Question

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Answer 1

Textbook Question

Chapter 15, Problem 4SP

For standard tuning, concert A is defined to have a frequency of 440 Hz. On a piano. A is five white keys above C, but 9 half steps above C counting both the white and black keys. (See fig. 15.22.) A full octave consists of 12 half steps (semitones). In equally tempered tuning, each half step has the ratio of 1.0595 above the preceding step. (This ratio is the 12th root of 2.0.)

a. What is the frequency of A-flat, one half step below A for equal temperament?

b. Working down, find the frequency of each succeeding half step until you get down to C. (Carry your computations to four figures, to avoid rounding errors. For each half step, divide by 1.0595.)

c. In just tuning, middle C has a frequency of 264 Hz. How does your result in part b compare to this value?

d. Working up, find the frequency of C above concert A in equal temperament. Is this frequency twice that obtained in part b for middle C?

(a)

Expert Solution

To determine

What is the frequency of A-flat.

Answer to Problem 4SP

The frequency of A flat is $415.3 Hz$ .

Explanation of Solution

Given info: The frequency of A is $440 Hz$

Write the formula to calculate the frequency of A flat.

$fA=(f1.0595)$

Here,

$fA$ is the frequency of A flat

f is the frequency of A

Substitute $440 Hz$ for f in the above equation to calculate $fA$ .

$fA=(440 Hz1.0595)=415.3 Hz$

Conclusion:

Therefore, the frequency of A flat is $415.3 Hz$ .

(b)

Expert Solution

To determine

Find the each succeeding half up to reach C.

Answer to Problem 4SP

The succeeding frequencies are $415.29 Hz$ , $391.97 Hz$ , $369.96 Hz$ , $349.18 Hz$ , $329.57 Hz$ , $311.06 Hz$ , $293.59 Hz$ , $277.10 Hz$ and $261.54 Hz$ .

Explanation of Solution

Given info: The frequency of A is $440 Hz$ and C is the 9 half steps below C.

Since C is the 9 half steps below A, to find the succeeding frequencies divide the frequency of A by 1.0595 to get the one step below of A and go on up to nine times. The 9^th frequency is corresponding to the frequency of C.

The succeeding frequencies are $415.29 Hz$ , $391.97 Hz$ , $369.96 Hz$ , $349.18 Hz$ , $329.57 Hz$ , $311.06 Hz$ , $293.59 Hz$ , $277.10 Hz$ and $261.54 Hz$ .

Conclusion:

Therefore, the succeeding frequencies are $415.29 Hz$ , $391.97 Hz$ , $369.96 Hz$ , $349.18 Hz$ , $329.57 Hz$ , $311.06 Hz$ , $293.59 Hz$ , $277.10 Hz$ and $261.54 Hz$ .

(c)

Expert Solution

To determine

Compare the given tuned frequency of C with part (b).

Answer to Problem 4SP

The difference in frequency is $2.5 Hz$ .

Explanation of Solution

Given Info: The tuned frequency of C is $264 Hz$ and frequency of C is $261.54 Hz$ .

Write the expression to calculate the difference between the tuned frequency of C with calculated.

$Δf=fC−fC′$

Here,

$Δf$ is the difference in frequency

$fC$ is the tuned frequency of C

$fC′$ is the calculate frequency of C

Substitute $264 Hz$ for $fC$ and $261.54 Hz$ for $fC′$ in the above equation to calculate $Δf$ .

$Δf=264 Hz−261.54 Hz=2.46 Hz≈2.5 Hz$

Conclusion:

Therefore, the difference in frequency is $2.5 Hz$ .

(d)

Expert Solution

To determine

The frequency of C above A and will this frequency is twice as compared with part (b).

Answer to Problem 4SP

The frequency of C is $523.3 Hz$ and which is almost twice as found before with a slight difference of $0.2$ .

Explanation of Solution

Given info: The frequency of A is $440 Hz$

In this case the C is five keys above A including white and black keys and three keys above without including black keys or halves. Therefore to find the frequency of C, the frequency of A is multiplied by the value 1.0595 by three times.

Therefore the frequency of C is $523.3 Hz$ and this frequency is almost twice as that of the frequency of C which is found to be $261.54 Hz$ with a slight difference of $0.2$ .

Conclusion:

Therefore, the frequency of C is $523.3 Hz$ and which is almost twice as found before with a slight difference of $0.2$ .

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Answer 2

Textbook Question

Chapter 15, Problem 4SP

For standard tuning, concert A is defined to have a frequency of 440 Hz. On a piano. A is five white keys above C, but 9 half steps above C counting both the white and black keys. (See fig. 15.22.) A full octave consists of 12 half steps (semitones). In equally tempered tuning, each half step has the ratio of 1.0595 above the preceding step. (This ratio is the 12th root of 2.0.)

a. What is the frequency of A-flat, one half step below A for equal temperament?

b. Working down, find the frequency of each succeeding half step until you get down to C. (Carry your computations to four figures, to avoid rounding errors. For each half step, divide by 1.0595.)

c. In just tuning, middle C has a frequency of 264 Hz. How does your result in part b compare to this value?

d. Working up, find the frequency of C above concert A in equal temperament. Is this frequency twice that obtained in part b for middle C?

(a)

Expert Solution

To determine

What is the frequency of A-flat.

Answer to Problem 4SP

The frequency of A flat is $415.3 Hz$ .

Explanation of Solution

Given info: The frequency of A is $440 Hz$

Write the formula to calculate the frequency of A flat.

$fA=(f1.0595)$

Here,

$fA$ is the frequency of A flat

f is the frequency of A

Substitute $440 Hz$ for f in the above equation to calculate $fA$ .

$fA=(440 Hz1.0595)=415.3 Hz$

Conclusion:

Therefore, the frequency of A flat is $415.3 Hz$ .

(b)

Expert Solution

To determine

Find the each succeeding half up to reach C.

Answer to Problem 4SP

The succeeding frequencies are $415.29 Hz$ , $391.97 Hz$ , $369.96 Hz$ , $349.18 Hz$ , $329.57 Hz$ , $311.06 Hz$ , $293.59 Hz$ , $277.10 Hz$ and $261.54 Hz$ .

Explanation of Solution

Given info: The frequency of A is $440 Hz$ and C is the 9 half steps below C.

Since C is the 9 half steps below A, to find the succeeding frequencies divide the frequency of A by 1.0595 to get the one step below of A and go on up to nine times. The 9^th frequency is corresponding to the frequency of C.

The succeeding frequencies are $415.29 Hz$ , $391.97 Hz$ , $369.96 Hz$ , $349.18 Hz$ , $329.57 Hz$ , $311.06 Hz$ , $293.59 Hz$ , $277.10 Hz$ and $261.54 Hz$ .

Conclusion:

Therefore, the succeeding frequencies are $415.29 Hz$ , $391.97 Hz$ , $369.96 Hz$ , $349.18 Hz$ , $329.57 Hz$ , $311.06 Hz$ , $293.59 Hz$ , $277.10 Hz$ and $261.54 Hz$ .

(c)

Expert Solution

To determine

Compare the given tuned frequency of C with part (b).

Answer to Problem 4SP

The difference in frequency is $2.5 Hz$ .

Explanation of Solution

Given Info: The tuned frequency of C is $264 Hz$ and frequency of C is $261.54 Hz$ .

Write the expression to calculate the difference between the tuned frequency of C with calculated.

$Δf=fC−fC′$

Here,

$Δf$ is the difference in frequency

$fC$ is the tuned frequency of C

$fC′$ is the calculate frequency of C

Substitute $264 Hz$ for $fC$ and $261.54 Hz$ for $fC′$ in the above equation to calculate $Δf$ .

$Δf=264 Hz−261.54 Hz=2.46 Hz≈2.5 Hz$

Conclusion:

Therefore, the difference in frequency is $2.5 Hz$ .

(d)

Expert Solution

To determine

The frequency of C above A and will this frequency is twice as compared with part (b).

Answer to Problem 4SP

The frequency of C is $523.3 Hz$ and which is almost twice as found before with a slight difference of $0.2$ .

Explanation of Solution

Given info: The frequency of A is $440 Hz$

In this case the C is five keys above A including white and black keys and three keys above without including black keys or halves. Therefore to find the frequency of C, the frequency of A is multiplied by the value 1.0595 by three times.

Therefore the frequency of C is $523.3 Hz$ and this frequency is almost twice as that of the frequency of C which is found to be $261.54 Hz$ with a slight difference of $0.2$ .

Conclusion:

Therefore, the frequency of C is $523.3 Hz$ and which is almost twice as found before with a slight difference of $0.2$ .

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Answer 3

Textbook Question

Chapter 15, Problem 4SP

For standard tuning, concert A is defined to have a frequency of 440 Hz. On a piano. A is five white keys above C, but 9 half steps above C counting both the white and black keys. (See fig. 15.22.) A full octave consists of 12 half steps (semitones). In equally tempered tuning, each half step has the ratio of 1.0595 above the preceding step. (This ratio is the 12th root of 2.0.)

a. What is the frequency of A-flat, one half step below A for equal temperament?

b. Working down, find the frequency of each succeeding half step until you get down to C. (Carry your computations to four figures, to avoid rounding errors. For each half step, divide by 1.0595.)

c. In just tuning, middle C has a frequency of 264 Hz. How does your result in part b compare to this value?

d. Working up, find the frequency of C above concert A in equal temperament. Is this frequency twice that obtained in part b for middle C?

(a)

Expert Solution

To determine

What is the frequency of A-flat.

Answer to Problem 4SP

The frequency of A flat is $415.3 Hz$ .

Explanation of Solution

Given info: The frequency of A is $440 Hz$

Write the formula to calculate the frequency of A flat.

$fA=(f1.0595)$

Here,

$fA$ is the frequency of A flat

f is the frequency of A

Substitute $440 Hz$ for f in the above equation to calculate $fA$ .

$fA=(440 Hz1.0595)=415.3 Hz$

Conclusion:

Therefore, the frequency of A flat is $415.3 Hz$ .

(b)

Expert Solution

To determine

Find the each succeeding half up to reach C.

Answer to Problem 4SP

The succeeding frequencies are $415.29 Hz$ , $391.97 Hz$ , $369.96 Hz$ , $349.18 Hz$ , $329.57 Hz$ , $311.06 Hz$ , $293.59 Hz$ , $277.10 Hz$ and $261.54 Hz$ .

Explanation of Solution

Given info: The frequency of A is $440 Hz$ and C is the 9 half steps below C.

Since C is the 9 half steps below A, to find the succeeding frequencies divide the frequency of A by 1.0595 to get the one step below of A and go on up to nine times. The 9^th frequency is corresponding to the frequency of C.

The succeeding frequencies are $415.29 Hz$ , $391.97 Hz$ , $369.96 Hz$ , $349.18 Hz$ , $329.57 Hz$ , $311.06 Hz$ , $293.59 Hz$ , $277.10 Hz$ and $261.54 Hz$ .

Conclusion:

Therefore, the succeeding frequencies are $415.29 Hz$ , $391.97 Hz$ , $369.96 Hz$ , $349.18 Hz$ , $329.57 Hz$ , $311.06 Hz$ , $293.59 Hz$ , $277.10 Hz$ and $261.54 Hz$ .

(c)

Expert Solution

To determine

Compare the given tuned frequency of C with part (b).

Answer to Problem 4SP

The difference in frequency is $2.5 Hz$ .

Explanation of Solution

Given Info: The tuned frequency of C is $264 Hz$ and frequency of C is $261.54 Hz$ .

Write the expression to calculate the difference between the tuned frequency of C with calculated.

$Δf=fC−fC′$

Here,

$Δf$ is the difference in frequency

$fC$ is the tuned frequency of C

$fC′$ is the calculate frequency of C

Substitute $264 Hz$ for $fC$ and $261.54 Hz$ for $fC′$ in the above equation to calculate $Δf$ .

$Δf=264 Hz−261.54 Hz=2.46 Hz≈2.5 Hz$

Conclusion:

Therefore, the difference in frequency is $2.5 Hz$ .

(d)

Expert Solution

To determine

The frequency of C above A and will this frequency is twice as compared with part (b).

Answer to Problem 4SP

The frequency of C is $523.3 Hz$ and which is almost twice as found before with a slight difference of $0.2$ .

Explanation of Solution

Given info: The frequency of A is $440 Hz$

In this case the C is five keys above A including white and black keys and three keys above without including black keys or halves. Therefore to find the frequency of C, the frequency of A is multiplied by the value 1.0595 by three times.

Therefore the frequency of C is $523.3 Hz$ and this frequency is almost twice as that of the frequency of C which is found to be $261.54 Hz$ with a slight difference of $0.2$ .

Conclusion:

Therefore, the frequency of C is $523.3 Hz$ and which is almost twice as found before with a slight difference of $0.2$ .

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Answer 4

Textbook Question

Chapter 15, Problem 4SP

For standard tuning, concert A is defined to have a frequency of 440 Hz. On a piano. A is five white keys above C, but 9 half steps above C counting both the white and black keys. (See fig. 15.22.) A full octave consists of 12 half steps (semitones). In equally tempered tuning, each half step has the ratio of 1.0595 above the preceding step. (This ratio is the 12th root of 2.0.)

a. What is the frequency of A-flat, one half step below A for equal temperament?

b. Working down, find the frequency of each succeeding half step until you get down to C. (Carry your computations to four figures, to avoid rounding errors. For each half step, divide by 1.0595.)

c. In just tuning, middle C has a frequency of 264 Hz. How does your result in part b compare to this value?

d. Working up, find the frequency of C above concert A in equal temperament. Is this frequency twice that obtained in part b for middle C?

(a)

Expert Solution

To determine

What is the frequency of A-flat.

Answer to Problem 4SP

The frequency of A flat is $415.3 Hz$ .

Explanation of Solution

Given info: The frequency of A is $440 Hz$

Write the formula to calculate the frequency of A flat.

$fA=(f1.0595)$

Here,

$fA$ is the frequency of A flat

f is the frequency of A

Substitute $440 Hz$ for f in the above equation to calculate $fA$ .

$fA=(440 Hz1.0595)=415.3 Hz$

Conclusion:

Therefore, the frequency of A flat is $415.3 Hz$ .

(b)

Expert Solution

To determine

Find the each succeeding half up to reach C.

Answer to Problem 4SP

The succeeding frequencies are $415.29 Hz$ , $391.97 Hz$ , $369.96 Hz$ , $349.18 Hz$ , $329.57 Hz$ , $311.06 Hz$ , $293.59 Hz$ , $277.10 Hz$ and $261.54 Hz$ .

Explanation of Solution

Given info: The frequency of A is $440 Hz$ and C is the 9 half steps below C.

Since C is the 9 half steps below A, to find the succeeding frequencies divide the frequency of A by 1.0595 to get the one step below of A and go on up to nine times. The 9^th frequency is corresponding to the frequency of C.

The succeeding frequencies are $415.29 Hz$ , $391.97 Hz$ , $369.96 Hz$ , $349.18 Hz$ , $329.57 Hz$ , $311.06 Hz$ , $293.59 Hz$ , $277.10 Hz$ and $261.54 Hz$ .

Conclusion:

Therefore, the succeeding frequencies are $415.29 Hz$ , $391.97 Hz$ , $369.96 Hz$ , $349.18 Hz$ , $329.57 Hz$ , $311.06 Hz$ , $293.59 Hz$ , $277.10 Hz$ and $261.54 Hz$ .

(c)

Expert Solution

To determine

Compare the given tuned frequency of C with part (b).

Answer to Problem 4SP

The difference in frequency is $2.5 Hz$ .

Explanation of Solution

Given Info: The tuned frequency of C is $264 Hz$ and frequency of C is $261.54 Hz$ .

Write the expression to calculate the difference between the tuned frequency of C with calculated.

$Δf=fC−fC′$

Here,

$Δf$ is the difference in frequency

$fC$ is the tuned frequency of C

$fC′$ is the calculate frequency of C

Substitute $264 Hz$ for $fC$ and $261.54 Hz$ for $fC′$ in the above equation to calculate $Δf$ .

$Δf=264 Hz−261.54 Hz=2.46 Hz≈2.5 Hz$

Conclusion:

Therefore, the difference in frequency is $2.5 Hz$ .

(d)

Expert Solution

To determine

The frequency of C above A and will this frequency is twice as compared with part (b).

Answer to Problem 4SP

The frequency of C is $523.3 Hz$ and which is almost twice as found before with a slight difference of $0.2$ .

Explanation of Solution

Given info: The frequency of A is $440 Hz$

In this case the C is five keys above A including white and black keys and three keys above without including black keys or halves. Therefore to find the frequency of C, the frequency of A is multiplied by the value 1.0595 by three times.

Therefore the frequency of C is $523.3 Hz$ and this frequency is almost twice as that of the frequency of C which is found to be $261.54 Hz$ with a slight difference of $0.2$ .

Conclusion:

Therefore, the frequency of C is $523.3 Hz$ and which is almost twice as found before with a slight difference of $0.2$ .

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Answer 5

Textbook Question

Answer 6

Textbook Question

Answer 7

(a)

Expert Solution

To determine

What is the frequency of A-flat.

Answer to Problem 4SP

The frequency of A flat is $415.3 Hz$ .

Explanation of Solution

Given info: The frequency of A is $440 Hz$

Write the formula to calculate the frequency of A flat.

$fA=(f1.0595)$

Here,

$fA$ is the frequency of A flat

f is the frequency of A

Substitute $440 Hz$ for f in the above equation to calculate $fA$ .

$fA=(440 Hz1.0595)=415.3 Hz$

Conclusion:

Therefore, the frequency of A flat is $415.3 Hz$ .

(b)

Expert Solution

To determine

Find the each succeeding half up to reach C.

Answer to Problem 4SP

The succeeding frequencies are $415.29 Hz$ , $391.97 Hz$ , $369.96 Hz$ , $349.18 Hz$ , $329.57 Hz$ , $311.06 Hz$ , $293.59 Hz$ , $277.10 Hz$ and $261.54 Hz$ .

Explanation of Solution

Given info: The frequency of A is $440 Hz$ and C is the 9 half steps below C.

Since C is the 9 half steps below A, to find the succeeding frequencies divide the frequency of A by 1.0595 to get the one step below of A and go on up to nine times. The 9^th frequency is corresponding to the frequency of C.

The succeeding frequencies are $415.29 Hz$ , $391.97 Hz$ , $369.96 Hz$ , $349.18 Hz$ , $329.57 Hz$ , $311.06 Hz$ , $293.59 Hz$ , $277.10 Hz$ and $261.54 Hz$ .

Conclusion:

Therefore, the succeeding frequencies are $415.29 Hz$ , $391.97 Hz$ , $369.96 Hz$ , $349.18 Hz$ , $329.57 Hz$ , $311.06 Hz$ , $293.59 Hz$ , $277.10 Hz$ and $261.54 Hz$ .

(c)

Expert Solution

To determine

Compare the given tuned frequency of C with part (b).

Answer to Problem 4SP

The difference in frequency is $2.5 Hz$ .

Explanation of Solution

Given Info: The tuned frequency of C is $264 Hz$ and frequency of C is $261.54 Hz$ .

Write the expression to calculate the difference between the tuned frequency of C with calculated.

$Δf=fC−fC′$

Here,

$Δf$ is the difference in frequency

$fC$ is the tuned frequency of C

$fC′$ is the calculate frequency of C

Substitute $264 Hz$ for $fC$ and $261.54 Hz$ for $fC′$ in the above equation to calculate $Δf$ .

$Δf=264 Hz−261.54 Hz=2.46 Hz≈2.5 Hz$

Conclusion:

Therefore, the difference in frequency is $2.5 Hz$ .

(d)

Expert Solution

To determine

The frequency of C above A and will this frequency is twice as compared with part (b).

Answer to Problem 4SP

The frequency of C is $523.3 Hz$ and which is almost twice as found before with a slight difference of $0.2$ .

Explanation of Solution

Given info: The frequency of A is $440 Hz$

In this case the C is five keys above A including white and black keys and three keys above without including black keys or halves. Therefore to find the frequency of C, the frequency of A is multiplied by the value 1.0595 by three times.

Therefore the frequency of C is $523.3 Hz$ and this frequency is almost twice as that of the frequency of C which is found to be $261.54 Hz$ with a slight difference of $0.2$ .

Conclusion:

Therefore, the frequency of C is $523.3 Hz$ and which is almost twice as found before with a slight difference of $0.2$ .

Answer 8

(a)

Expert Solution

To determine

What is the frequency of A-flat.

Answer to Problem 4SP

The frequency of A flat is $415.3 Hz$ .

Explanation of Solution

Given info: The frequency of A is $440 Hz$

Write the formula to calculate the frequency of A flat.

$fA=(f1.0595)$

Here,

$fA$ is the frequency of A flat

f is the frequency of A

Substitute $440 Hz$ for f in the above equation to calculate $fA$ .

$fA=(440 Hz1.0595)=415.3 Hz$

Conclusion:

Therefore, the frequency of A flat is $415.3 Hz$ .

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

To determine

What is the frequency of A-flat.

Answer 13

Answer to Problem 4SP

The frequency of A flat is $415.3 Hz$ .

Answer 14

Explanation of Solution

Given info: The frequency of A is $440 Hz$

Write the formula to calculate the frequency of A flat.

$fA=(f1.0595)$

Here,

$fA$ is the frequency of A flat

f is the frequency of A

Substitute $440 Hz$ for f in the above equation to calculate $fA$ .

$fA=(440 Hz1.0595)=415.3 Hz$

Conclusion:

Therefore, the frequency of A flat is $415.3 Hz$ .

Answer 15

(b)

Expert Solution

To determine

Find the each succeeding half up to reach C.

Answer to Problem 4SP

The succeeding frequencies are $415.29 Hz$ , $391.97 Hz$ , $369.96 Hz$ , $349.18 Hz$ , $329.57 Hz$ , $311.06 Hz$ , $293.59 Hz$ , $277.10 Hz$ and $261.54 Hz$ .

Explanation of Solution

Given info: The frequency of A is $440 Hz$ and C is the 9 half steps below C.

Since C is the 9 half steps below A, to find the succeeding frequencies divide the frequency of A by 1.0595 to get the one step below of A and go on up to nine times. The 9^th frequency is corresponding to the frequency of C.

The succeeding frequencies are $415.29 Hz$ , $391.97 Hz$ , $369.96 Hz$ , $349.18 Hz$ , $329.57 Hz$ , $311.06 Hz$ , $293.59 Hz$ , $277.10 Hz$ and $261.54 Hz$ .

Conclusion:

Therefore, the succeeding frequencies are $415.29 Hz$ , $391.97 Hz$ , $369.96 Hz$ , $349.18 Hz$ , $329.57 Hz$ , $311.06 Hz$ , $293.59 Hz$ , $277.10 Hz$ and $261.54 Hz$ .

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

To determine

Find the each succeeding half up to reach C.

Answer 20

Answer to Problem 4SP

The succeeding frequencies are $415.29 Hz$ , $391.97 Hz$ , $369.96 Hz$ , $349.18 Hz$ , $329.57 Hz$ , $311.06 Hz$ , $293.59 Hz$ , $277.10 Hz$ and $261.54 Hz$ .

Answer 21

Explanation of Solution

Given info: The frequency of A is $440 Hz$ and C is the 9 half steps below C.

Since C is the 9 half steps below A, to find the succeeding frequencies divide the frequency of A by 1.0595 to get the one step below of A and go on up to nine times. The 9^th frequency is corresponding to the frequency of C.

The succeeding frequencies are $415.29 Hz$ , $391.97 Hz$ , $369.96 Hz$ , $349.18 Hz$ , $329.57 Hz$ , $311.06 Hz$ , $293.59 Hz$ , $277.10 Hz$ and $261.54 Hz$ .

Conclusion:

Therefore, the succeeding frequencies are $415.29 Hz$ , $391.97 Hz$ , $369.96 Hz$ , $349.18 Hz$ , $329.57 Hz$ , $311.06 Hz$ , $293.59 Hz$ , $277.10 Hz$ and $261.54 Hz$ .

Answer 22

(c)

Expert Solution

To determine

Compare the given tuned frequency of C with part (b).

Answer to Problem 4SP

The difference in frequency is $2.5 Hz$ .

Explanation of Solution

Given Info: The tuned frequency of C is $264 Hz$ and frequency of C is $261.54 Hz$ .

Write the expression to calculate the difference between the tuned frequency of C with calculated.

$Δf=fC−fC′$

Here,

$Δf$ is the difference in frequency

$fC$ is the tuned frequency of C

$fC′$ is the calculate frequency of C

Substitute $264 Hz$ for $fC$ and $261.54 Hz$ for $fC′$ in the above equation to calculate $Δf$ .

$Δf=264 Hz−261.54 Hz=2.46 Hz≈2.5 Hz$

Conclusion:

Therefore, the difference in frequency is $2.5 Hz$ .

Answer 23

(c)

Expert Solution

Answer 24

(c)

Expert Solution

Answer 25

Expert Solution

Answer 26

To determine

Compare the given tuned frequency of C with part (b).

Answer 27

Answer to Problem 4SP

The difference in frequency is $2.5 Hz$ .

Answer 28

Explanation of Solution

Given Info: The tuned frequency of C is $264 Hz$ and frequency of C is $261.54 Hz$ .

Write the expression to calculate the difference between the tuned frequency of C with calculated.

$Δf=fC−fC′$

Here,

$Δf$ is the difference in frequency

$fC$ is the tuned frequency of C

$fC′$ is the calculate frequency of C

Substitute $264 Hz$ for $fC$ and $261.54 Hz$ for $fC′$ in the above equation to calculate $Δf$ .

$Δf=264 Hz−261.54 Hz=2.46 Hz≈2.5 Hz$

Conclusion:

Therefore, the difference in frequency is $2.5 Hz$ .

Answer 29

(d)

Expert Solution

To determine

The frequency of C above A and will this frequency is twice as compared with part (b).

Answer to Problem 4SP

The frequency of C is $523.3 Hz$ and which is almost twice as found before with a slight difference of $0.2$ .

Explanation of Solution

Given info: The frequency of A is $440 Hz$

In this case the C is five keys above A including white and black keys and three keys above without including black keys or halves. Therefore to find the frequency of C, the frequency of A is multiplied by the value 1.0595 by three times.

Therefore the frequency of C is $523.3 Hz$ and this frequency is almost twice as that of the frequency of C which is found to be $261.54 Hz$ with a slight difference of $0.2$ .

Conclusion:

Therefore, the frequency of C is $523.3 Hz$ and which is almost twice as found before with a slight difference of $0.2$ .

Answer 30

(d)

Expert Solution

Answer 31

(d)

Expert Solution

Answer 32

Expert Solution

Answer 33

To determine

The frequency of C above A and will this frequency is twice as compared with part (b).

Answer 34

Answer to Problem 4SP

The frequency of C is $523.3 Hz$ and which is almost twice as found before with a slight difference of $0.2$ .

Answer 35

Explanation of Solution

Given info: The frequency of A is $440 Hz$

In this case the C is five keys above A including white and black keys and three keys above without including black keys or halves. Therefore to find the frequency of C, the frequency of A is multiplied by the value 1.0595 by three times.

Therefore the frequency of C is $523.3 Hz$ and this frequency is almost twice as that of the frequency of C which is found to be $261.54 Hz$ with a slight difference of $0.2$ .