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Question

String A has a length L, radius of cross-section r, density of material ρ and is under tension T. String B has all these quantities double those of string A. If υA and υB are corresponding fundamental frequencies of the vibrating string, then

A
υA=2υB
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B
υA=4υB
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C
υB=4υA
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D
υA=υB
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Solution

The correct option is B υA=4υB
Fundamental frequency of a string is
υ=12LTπr2ρ=12LrIπρ

where the symbols have their usual meanings.
υAυB=(LBLA)(rBrA)(TATBρBρA)1/2
Substituting the given values, we get
υAυB=(2LL)(2rr)(T2T2ρρ)1/2=4,υA=4υB

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