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 $υ_{A}$ and $υ_{B}$ are corresponding fundamental frequencies of the vibrating string, then

υ_{A} = 2 υ_{B}

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υ_{A} = 4 υ_{B}

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υ_{B} = 4 υ_{A}

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υ_{A} = υ_{B}

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Solution

The correct option is B $υ_{A} = 4 υ_{B}$
Fundamental frequency of a string is
$υ = \frac{1}{2 L} \sqrt{\frac{T}{π r^{2} ρ}} = \frac{1}{2 L r} \sqrt{\frac{I}{π ρ}}$

where the symbols have their usual meanings.
$∴ \frac{υ_{A}}{υ_{B}} = (\frac{L_{B}}{L_{A}}) (\frac{r_{B}}{r_{A}}) {(\frac{T_{A}}{T_{B}} \frac{ρ_{B}}{ρ_{A}})}^{1 / 2}$
Substituting the given values, we get
$\frac{υ_{A}}{υ_{B}} = (\frac{2 L}{L}) (\frac{2 r}{r}) {(\frac{T}{2 T} \frac{2 ρ}{ρ})}^{1 / 2} = 4, υ_{A} = 4 υ_{B}$

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