problem17_82

University Physics with Modern Physics with Mastering Physics (11th Edition)

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17.82: (a) We can use differentials to find the frequency change because all length changes are small percents . Let m be the mass of the wire m FL L m F μ F v = = = ) ( mL F L m FL v f L v f 2 1 2 ) l fundamenta ( 2 and = = λ = = λ λ = L L F f (only L changes due to heating) L L L mL F f f mL F mL F = - - 2 1 ) ( ) )( ( 2 1 2 1 2 1 2 1 2 Hz 15 . 0 ) Hz 440 )( C 440 )( C 40 )( ) C ( 10 7 . 1 ( 2 1 ) ( 2 1 1 5 = ° ° ° × = - - f T α f The frequency decreases since the length increases (b) m FL μ F v = = 2 2 ) ( ) ( 2 1 2 1 T α L L m FL L m
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Unformatted text preview: F m FL v v ∆ = ∆ = ∆ = ∆-% 034 . 10 4 . 3 ) C 40 )( ) C ( 10 7 . 1 ( 2 1 4 1 5 = × = ° ° × =---(c) T α L L L L L L ∆ = = = → ∆ = λ ∆ → = λ ∆ ∆ λ λ ∆ 2 2 2 2 : % 068 . 10 8 . 6 ) C 40 )( C 10 7 . 1 ( 4 1 5 = × = ° × = λ λ ∆--- it increases...
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This document was uploaded on 02/05/2008.

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