Problem 12.43 Part A When a player\'s finger presses a guitar string down onto a
ID: 1658146 • Letter: P
Question
Problem 12.43 Part A When a player's finger presses a guitar string down onto a fret, the length of the vibrating portion of the string is shortened, thereby increasing the string's fundamental frequency (Figure 1). The string's tension and mass per unit length remain unchanged. If the unfingered length of the string is L-65.0 cm , determine the positions of the first six frets, if each fret raises the pitch of the fundamental by one musical note compared to the neighboring fret. On the equally tempered chromatic scale, the ratio of frequencies of neighboring notes is Determine the position of the first fret. Express your answer to one decimal place and include the appropriate units Value Units 1/12 Submit My Answers Give Up Part B Determine the position of the second fret. Express your answer to one decimal place and include the appropriate units Figure 1 of 1 | Value Units Submit My Answers Give Up Part C Determine the position of the third fret.Explanation / Answer
As we know that the wavelength of the fundamental frequency is twice the string's length (the fret and the bridge are like the first and second nodes of a sine wave).
So, we can write -
= 2L
And the wavelength is related to the frequency by v=f·, where v is the speed of sound.
therefore -
2L = v / f
L = v / (2f)
Now given that -
L0 = 65 cm and the ratio of that frequency to the frequency of the first fret note = 2^(1/12).
From this, you can find the length of the string for the first fret -
L1 / L0 = [v/(2f1)] / [v/(2f0)] = f0 / f1
=> L1 = 2^(1/12) * L0 = 0.167*65 = 10.83 cm
So, x1 = 65 - 10.83 = 54.17 cm
Part B:
So, x2 = 65 - 1.80 = 63.2 cm
L2 = 2^(1/12) * L1 = 0.167*10.83 = 1.80
Part C:
L3 = 2^(1/12) * L2 = 0.167*1.80 = 0.30
So, x3 = 65 - 0.30 = 64.70 cm
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