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A musical interval of an octave corresponds to a factor of
Chapter 15, Problem 21DQ(choose chapter or problem)
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?
Questions & Answers
QUESTION:
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?
ANSWER: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.