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A wire has a length of 114 cm between two fixed ends. Where should two bridges be placed to divide the wire into three segments whose fundamental frequencies are in the ratio 1: 3:4 ? |
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Answer» Solution :In case of a given wire under specific TENSION, fundamental frequency of vibration `f prop (1//L)`. So for having fundamental frequencies in the ratio of ` 1: 3 : 4`, the vibrating length should be in the ratio `1 : (1/3) : (1/4)`, i.e., `L_1 : L_2 : L_3 : : 12 : 4 : 3` COMMON factor is x, `12 x + 4x + 3X = 114 =` length of the string `19x = 114` or x = 6` `RARR L_1 = 12 xx 6 = 72 cm,L_2 cm,L_2 = 4 xx 6 = 24 cm, L_3 = 3 xx 6 = 18 cm` |
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