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# answer3 - Dimensional analysis problems Math 442.500 1 Why...

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Dimensional analysis problems Math 442.500 1. Why do stringed musical instruments have strengths of different lengths and thicknesses? Assume that the fundamental frequency ω of vibration of a string depends on its length l , mass per unit length μ , and tension (force) F on the string. Prove that ω must be proportional to p F/μ l . Answer: The physical quantities involved are frequency ω , mass per unit length μ , length l , and tension F . The dimensions of these are: [ ω ] = T - 1 , [ μ ] = ML - 1 , [ l ] = L , and [ F ] = MLT - 2 , where the dimensions of frequency and force are given in the class notes. A product of these is Π = ω a μ b l c F d , which has dimensions [Π] = T - a ( ML - 1 ) b L c ( MLT - 2 ) d = T - a - 2 d M b + d L - b + c + d . This will be dimensionless if and only if - a - 2 d = 0 b + d = 0 - b + c + d = 0 . The augmented matrix for this system is - 1 0 0 - 2 . . . 0 0 1 0 1 . . . 0 0 - 1 1 1 . . . 0 , which row reduces to 1 0 0 2 . . . 0 0 1 0 1 . . . 0 0 0 1 2 . . . 0 . The interpretation of this is that d is arbitrary, c = - 2 d , b = - d , and a = - 2 d . Thus the only possibility for a dimensionless Π is Π = ( ω - 2 μ - 1 l - 2 F ) d . The arbitrary power d may be ignored, since a function of ( ω - 2 μ - 1 l - 2 F ) d is also a function of ω - 2 μ - 1 l - 2 F . Thus a physical law relating these quantities must be of the form f ( ω - 2 μ - 1 l - 2 F ) = 0 , 1

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which we may generically solve for Π to get ω - 2 μ - 1 l - 2 F = k, for some dimensionless constant k . Thus ω = s kF μl 2 = k p F/μ l , as desired. 2. We now want to include frictional effects and the initial angle in analyzing pendulums. As in the lecture notes, the length of the pendulum is l and its mass is m . (a) Suppose that the frictional force is due primarily to air and is pro- portional to v 2 with constant of proportionality k . Let τ be the time required for the pendulum to reach half its initial amplitude θ . Determine the dimensions of k . Show that τ = s l g G θ, kl m for some function G . Answer: The quantities involved are τ , length l , mass m , initial amplitude θ , acceleration due to gravity g , and the drag constant k . The dimensions of all but the last quantity are clear. If drag force is k times v 2 , then the dimensions of that equation are MLT - 2 = [ k ] L 2 T - 2 , so k must have dimensions of ML - 1 . A product Π of the six quantities involved is Π = τ a l b m c θ d g e k f , which has dimensions [Π] = T a L b M
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