COLLEGE PHYSICS,VOLUME 1
COLLEGE PHYSICS,VOLUME 1
2nd Edition
ISBN: 9781319115104
Author: Freedman
Publisher: MAC HIGHER
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Chapter 13, Problem 104QAP
To determine

(a)

Show that the ratio of the frequency of any semitone to the frequency of the note just below it is 21/12

Expert Solution
Check Mark

Answer to Problem 104QAP

The ratio semitone to the frequency of the note just below is 21/12

Explanation of Solution

Given:

The 12 semitones of octave are as follow.

  C,C*/Db,D,D#/Eb,E,F,F#/Gb,G,G#/Ab,A,A#/Bb,B,C

Introduction:An octave or perfect octave is the interval between one musical pitch and another with double its frequency.

  fC'fC=2where,fC=firstfrequencyofactavefC'=secondfrequencyofoctave

Further octaves of a note occur at times the frequency of that note (where n is an integer), such as 2, 4, 8, 16, 32, 64 etc. and the reciprocal of that series.

The octave is divide into 12 semitone of equal ratio r.

  (f C fB)=(fBf A * )=(f A * fA)=(fAf G * )=(f G * fG)=(fGf F * )=(f F * fF)=(fFfE)=(fEf D * )=(fDfD)=(fDf C * )=(f C * fC)=r

Calculation:

  fCfC=(f C fB)(fBf A * )(f A * fA)(fAf G * )(f G * fG)(fGf F * )(f F * fF)(fFfE)(fEf D * )(fDfD)(fDf C * )(f C * fC)fCfC=r122=r12r=21/12

Conclusion:

The ratio semitone to the frequency of the note just below is 21/12

To determine

(b)

The frequency of F# in the tempered scale, If A is 440 Hz.

Expert Solution
Check Mark

Answer to Problem 104QAP

The frequency of F# is 370 Hz

Explanation of Solution

Formula used:

The ratio of the frequency of any semitone to the frequency of the note just below it is 21/12

  (fAfG*)=(fG*fG)=(fGfF*)=2112

Calculation:

  fAfF*=(fAf G * )(f G * fG)(fGf F * )=(2 1 12)3=214fF*=(214)fA=(214)(440Hz)=370Hz

Conclusion:

The frequency of F# is 370 Hz.

To determine

The value by which tension should change so that the string tuned from Bb to B.

Expert Solution
Check Mark

Answer to Problem 104QAP

The tension should be increased by a factor of 1.12

Explanation of Solution

Given:

  fBfBb=2112

Formula used:

  f=vλ=Fμwhere,f=requencyv=velocityF=tensionμ=linearmassdensity

Calculation:

  fBfBb=( v B λ)( v B b λ)=( F B μ )( F B 6 μ )=FBF B t =2112FBFBb=(2 1 12)2=216FB=(216)FBb=1.12FBb

Conclusion:

The tension should be increased by a factor of 1.12 so a string on a musical instrument is tuned from Bb to B

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Chapter 13 Solutions

COLLEGE PHYSICS,VOLUME 1

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