Physics of Everyday Phenomena
Physics of Everyday Phenomena
9th Edition
ISBN: 9781259894008
Author: W. Thomas Griffith, Juliet Brosing Professor
Publisher: McGraw-Hill Education
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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
Check Mark
To determine

What is the frequency of A-flat.

Answer to Problem 4SP

The frequency of A flat is 415.3Hz.

Explanation of Solution

Given info: The frequency of A is 440Hz

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

fA=(440Hz1.0595)=415.3Hz

Conclusion:

Therefore, the frequency of A flat is 415.3Hz.

(b)

Expert Solution
Check Mark
To determine

Find the each succeeding half up to reach C.

Answer to Problem 4SP

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

Explanation of Solution

Given info: The frequency of A is 440Hz 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 9th frequency is corresponding to the frequency of C.

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

Conclusion:

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

(c)

Expert Solution
Check Mark
To determine

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

Answer to Problem 4SP

The difference in frequency is 2.5Hz.

Explanation of Solution

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

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

Δf=fCfC

Here,

Δf is the difference in frequency

fC is the tuned frequency of C

fC is the calculate frequency of C

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

Δf=264Hz261.54Hz=2.46Hz2.5Hz

Conclusion:

Therefore, the difference in frequency is 2.5Hz.

(d)

Expert Solution
Check Mark
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 523.3Hz 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 440Hz

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

Conclusion:

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

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Chapter 15 Solutions

Physics of Everyday Phenomena

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