A wire of density $9 \times 10^{-3} \,kg\, cm ^{-3}$ is stretched between two clamps $1\, m$ apart. The resulting strain in the wire is $4.9 \times 10^{-4}$. The lowest frequency of the transverse vibrations in the wire is......$HZ$
(Young's modulus of wire $Y =9 \times 10^{10}\, Nm ^{-2}$ ), (to the nearest integer),
JEE MAIN 2020, Diffcult
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$\rho_{\text {wire }}=9 \times 10^{-3} \frac{ kg }{ cm ^{3}}=\frac{9 \times 10^{-3}}{10^{-6}} kg / m ^{3}$
$=9000 kg / m ^{2}$
$( A = CSA$ of wire $)$
$\left( Y =9 \times 10^{10} Nm ^{2}\right)$
$\left(\right.$ Strain $\left.=4.9 \times 10^{-4}\right)$
$\Rightarrow L =1 m =\frac{\lambda}{2} \Rightarrow \lambda=2 m$
$\Rightarrow v=f \lambda \Rightarrow \sqrt{\frac{T}{\mu}}=f \lambda$
Where $Y=\frac{T / A}{\text { strain }} \Rightarrow T=Y . A .$ strain
$\Rightarrow \sqrt{\frac{ Y . A . strain }{ m / L }}= f \times 2 \Rightarrow \sqrt{\frac{ Y . A I . \text { strain }}{ M }}= f \times 2$
$\Rightarrow \sqrt{\frac{ Y \times V \times \text { strain }}{ M }}= f \times 2 \Rightarrow \sqrt{\frac{ Y \times \text { strain }}{\rho}}= f \times 2$
$f =\frac{1}{2} \sqrt{\frac{ Y \times \text { strain }}{\rho}}=\frac{1}{2} \sqrt{\frac{9 \times 10^{10} \times 4.9 \times 10^{-4}}{9000}}$
$f=\frac{1}{2} \sqrt{\frac{9 \times 10^{3}}{9} \times 4.9}=\frac{1}{2} \sqrt{4900}=\frac{70}{2}=35 Hz$
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