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Two wires are together end to end . The wires are made of the same material , but the diameter of one is twice that of the other . They are subjected to a tension of 4.60 N. The thin wire has a length of 40.0 cm and a linear mass density of 2.00 g//m. The combination is fixed at both ends and vibratedin such a waythat two antinodes are present , with the node between them being precisely at the weld . (a) What is the frequency of vibration ? (b) Find the length of the thick wire . |
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Answer» Solution :The mass per volume density , the tension , and the frequency must be the same for the TWO wires . The linear density , wave SPEED , wavelength and node - to node distance for it . a. Since the first node is at the weld , the wavelength in the thin wire is ` lmbda = 2 L = 80.0 cm` The frequency and tension are the same in both sections , so ` f = (v)/( lambda) = (1)/( 2L) sqrt ((T)/( MU)) = ( 1)/( 2(0.400 m)) sqrt (( 4.60 N)/( 0.00200 kg//m)) = 59.9 Hz` B. Since the thick wire is twice the diameter , it will have four times the cross - sectional area , and a linear density ` mu'` that is four times that of the thin wire . `mu' = 4 ( 2.00 g//m) = 0.00800 kg//m`. `L' = (lambda')/( 2) = (v')/( 2 f) = (1)/( 2 f) sqrt ((T)/( mu'))` `= (1)/( 2( 59.9 Hz)) sqrt(( 4.60 N)/( 0.00800 kg//m)) = 20.0 cm` Note that the thick wire is half the length of the thin wire . We could have reasoned the answer by noting that the wave speed on the thick wire is half as large , so the wavelength should be half as large for the same frequency . |
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