sol2 - PHYS 218 - SOLUTION TO ASSIGNMENT 2 09 Feb 2007 By...

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Unformatted text preview: PHYS 218 - SOLUTION TO ASSIGNMENT 2 09 Feb 2007 By Chung Koo Kim e-mail : [email protected] 1. A musician tuning instrument The velocity of waves along a string is given by v = s T μ = fλ (a) As both ends of the string are fixed, the wavelength is also constant. So for the musician to increase the frequency for tuning, she should increase the tension. (b) With everything else kept constant, f ∝ √ T or T ∝ f 2 tells us that τ c τ i = f c fi ! 2 = 440 435 2 ' 1 . 023 2. Piano strings (a) For the fundamental frequency to be 261.6Hz, v = fλ = f (2 L ) = s T μ = s SA ρAL/L = s S ρ gives L = 1 2 f s S ρ ' . 604( m ) (b) Tension in the spring : S × A = 200 π ' 628( N ) Weight of a 150lb person : 150 × . 453 × 9 . 8 ' 665 . 91( N ) So the tension applied in the piano string is slightly less than, or almost as big as the weight of a 150lb person. Consequently, the total tension on the piano strings, of total number 88 × 3 = 264 (even more than this in practice) is the weight of mass of about 18ton! This force is enough to distort and eventually break the wooden frame of the piano. (c) From the first equation, the tension becomes T = (2 fL ) 2 μ = 4 f 2 L 2 ρA So the range of tension is between 6.934(N) and 1 . 607 × 10 5 (N). (d) Since the cross-section area is A = πD 2 / 4, the equation in (c) tells us that D = 1 fL s T πρ giving the range of diameter between 9.519mm and 62.54giving the range of diameter between 9....
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This note was uploaded on 03/13/2008 for the course PHYS 2218 taught by Professor Wittich,p during the Spring '08 term at Cornell.

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sol2 - PHYS 218 - SOLUTION TO ASSIGNMENT 2 09 Feb 2007 By...

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