Mass per Unit Length of Violin Strings

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The discussion focuses on determining the mass per unit length of violin strings when each string is tuned to a frequency 1.5 times that of its neighbor under the same tension. The fundamental frequency of the lowest string is expressed as f1 = v1/(2L), leading to the relationship fn = 1.5^(n-1)f1 for higher strings. The velocity of the standing wave is linked to tension and mass per unit length, resulting in the formula μn = μ1 / 1.5^(2(n-1)). This calculation has been verified against a textbook answer, confirming its accuracy. The problem is effectively solved with the derived relationships.
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[SOLVED] Mass per Unit Length of Violin Strings

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


f = v/(2L)
v = \sqrt{T / \mu }

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,

1.5^{n - 1}v_1 = v_n

Now, since v_i = \sqrt{T/\mu_i}, then

\mu_n = \frac{\mu_1}{1.5^{2(n-1)}}

Is that right?
 
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I have checked the answer with the book. It coincides. Thanks anyways.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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