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[SOLVED] Mass per Unit Length of Violin Strings
Each string on a violin is tuned to a frequency 1.5 times that of its neighbor. If all the strings are to be placed under the same tension, what must be the mass per unit length of each string relative to that of the lowest string?
[tex]f = v/(2L)[/tex]
[tex]v = \sqrt{T / \mu }[/tex]
Suppose the lowest string is tuned at the fundamental frequency f1 = v1/(2L), where v1 is the velocity of the standing wave on the lowest string and L being the length of the string. For n > 1, fn = 1.5n - 1f1 = vn/(2L). So,
[tex]1.5^{n - 1}v_1 = v_n[/tex]
Now, since [itex]v_i = \sqrt{T/\mu_i}[/itex], then
[tex]\mu_n = \frac{\mu_1}{1.5^{2(n-1)}}[/tex]
Is that right?
Homework Statement
Each string on a violin is tuned to a frequency 1.5 times that of its neighbor. If all the strings are to be placed under the same tension, what must be the mass per unit length of each string relative to that of the lowest string?
Homework Equations
[tex]f = v/(2L)[/tex]
[tex]v = \sqrt{T / \mu }[/tex]
The Attempt at a Solution
Suppose the lowest string is tuned at the fundamental frequency f1 = v1/(2L), where v1 is the velocity of the standing wave on the lowest string and L being the length of the string. For n > 1, fn = 1.5n - 1f1 = vn/(2L). So,
[tex]1.5^{n - 1}v_1 = v_n[/tex]
Now, since [itex]v_i = \sqrt{T/\mu_i}[/itex], then
[tex]\mu_n = \frac{\mu_1}{1.5^{2(n-1)}}[/tex]
Is that right?
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