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A wire whose linear density is $5 \times 10^{- 3} kg / m$ stretched between two points with a tension 450N. The wire resonates at a frequency of 420Hz. The next higher frequency at which the same wire resonates is 490Hz the length of the wire is ….m(round off to 1 decimal place)

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answer is 2.1.

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Detailed Solution

From the given data

fundamental frequency = 490Hz - 420Hz = 70Hz

Fundamental frequency in a string is

$f = \frac{1}{2 l} \sqrt{\frac{T}{μ}}$

$\Rightarrow 70 = \frac{1}{2 l} \sqrt{\frac{450}{5 \times 10^{- 3}}}$

$\Rightarrow l = \frac{300}{140} = 2.1$

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