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),

\(=9000 kg / m ^{2}\)

\(( A = CSA\) 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

\(\Rightarrow \sqrt{\frac{ Y . A . strain }{ m / L }}= f \times 2 \Rightarrow \sqrt{\frac{ Y . A I . \text { strain }}{ M }}= f \times 2\)

\(\Rightarrow \sqrt{\frac{ Y \times V \times \text { strain }}{ M }}= f \times 2 \Rightarrow \sqrt{\frac{ Y \times \text { strain }}{\rho}}= f \times 2\)

\(f =\frac{1}{2} \sqrt{\frac{ Y \times \text { strain }}{\rho}}=\frac{1}{2} \sqrt{\frac{9 \times 10^{10} \times 4.9 \times 10^{-4}}{9000}}\)

\(f=\frac{1}{2} \sqrt{\frac{9 \times 10^{3}}{9} \times 4.9}=\frac{1}{2} \sqrt{4900}=\frac{70}{2}=35 Hz\)