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  • A wire of density <img alt="8 \times 10^{3} \mathrm{~kg} / \mathrm{m}^{3}" src="https://entrancecorner.oncodecogs.com/gif.latex?8%20%5Ctimes%2010%5E%7B3%7D%20%5Cmathrm%7B%7Ekg%7D%20/%20%5Cmathrm%7B

A wire of density 8 \times 10^{3} \mathrm{~kg} / \mathrm{m}^{3} is stretched between two clamps 0.5 \mathrm{~m} apart. The extension developed in the wire is 3.2 \times 10^{-4} \mathrm{~m}.If \mathrm{Y}=8 \times 10^{10} \mathrm{~N} / \mathrm{m}^{2}, the fundamental frequency of vibration in the wire will be___________ \mathrm{Hz}.

Option: 1

80


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

best_answer

f=\frac{1}{2 \ell} \sqrt{\frac{T}{\mu}}
f=\frac{1}{2 \ell} \sqrt{\frac{Y A \Delta \ell}{\ell \mu}}
f=\frac{1}{2 \times 0.5} \sqrt{\frac{8 \times 10^{10} \times 3.2 \times 10^{-4}}{8 \times 10^{3} \times 0.5}}
f=\frac{1}{1} \sqrt{6400}
f=80 \mathrm{~Hz}

Posted by

chirag

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