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A musical interval of an octave corresponds to a factor of

University Physics | 13th Edition | ISBN: 9780321675460 | Authors: Hugh D. Young, Roger A. Freedman

Problem 21DQ Chapter 15

University Physics | 13th Edition

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University Physics | 13th Edition | ISBN: 9780321675460 | Authors: Hugh D. Young, Roger A. Freedman

University Physics | 13th Edition

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Problem 21DQ

A musical interval of an ?octave corresponds to a factor of 2 in frequency. By what factor must the tension in a guitar or violin string be increased to raise its pitch one octave? To raise it two octaves? Explain your reasoning. Is there any danger in attempting these changes in pitch?

Step-by-Step Solution:

Solution 21DQ Step 1: Suppose, the initial frequency is, 1 T 0= 2L Where, T - tension on the string - Linear mass density of the string L - length of the string If we are raising the pitch to one octave, then, the frequency will0be, 2 1 T So, we can write,20= 2L Where, T’ - New tension on the string Dividing both equations, 2 0 0= 1 T/1 T 2L 2L T 2 = T Squaring both sides, we get, 4 = T’ / T Or, T’ = 4T So, we have to apply a tension 4 times higher than the initial tension to raise the pitch to one octave. Step 2: For, two octaves, the frequency will be, 4 0 Therefore, we can write, 1 T " 4 0 2L Where T ” - tension to raise the frequency to two octaves Dividing the initial equation for frequen0y with this. 4 0/ 0 = 1 T "/1 T 2L 2L T " 4 = T Squaring both sides, we get, 16 = T ” / T Or, T ” = 16T So, we have to apply a tension 16 times higher than the initial tension to raise the pitch to two octaves.

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Chapter 15, Problem 21DQ is Solved
Textbook: University Physics
Edition: 13
Author: Hugh D. Young, Roger A. Freedman
ISBN: 9780321675460

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