This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.View Full Document
Unformatted text preview: 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 = b + d =- b + c + d = . The augmented matrix for this system is - 1- 2 . . . 1 1 . . .- 1 1 1 . . . , which row reduces to 1 2 . . . 1 1 . . . 1 2 . . . . 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 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 ....
View Full Document
This note was uploaded on 03/08/2010 for the course MATH 442 taught by Professor Howard during the Spring '08 term at Texas A&M.
- Spring '08