A wire of density 9$\times$10³ kg/m³ is stretched between two clamps 1 m apart and is subjected to an extension of 4.9$\times$10^-4 m. What will be the lowest frequency of transverse vibration in the wire (in Hz)? (Young's modulus of material = 9$\times$10¹⁰ N/m²).

Let A be the area of cross-section of wire and T be the tension applied. 

The lowest frequency of transverse vibration is given by ${\mathrm{n}}_0=\frac{1}{2\mathrm{l}}\sqrt{\left(\frac{\mathrm{T}}{\mathrm{\mu }}\right)}$

where $\mu =$ mass per unit length of wire
= Volume of unit length$\times$density

$=\mathrm{A}\times 1\times \mathrm{\rho }=\mathrm{A\rho }$

we know that Young's modulus $\mathrm{Y}=\frac{\text{ stress }}{\text{ strain }}=\frac{\mathrm{T}/\mathrm{A}}{\mathrm{\Delta l}/\mathrm{l}}\Rightarrow \mathrm{T}=\frac{\mathrm{YA\Delta l}}{\mathrm{l}}$

      ${\mathrm{n}}_{\mathrm{o}}=\frac{1}{2\mathrm{l}}\sqrt{\left(\frac{\mathrm{YA\Delta l}}{\mathrm{lA\rho }}\right)}=\frac{1}{2\mathrm{l}}\sqrt{\left(\frac{\mathrm{Y\Delta l}}{\mathrm{l\rho }}\right)}$

Substituting the given values

    ${\mathrm{n}}_0=\frac{1}{2\times 1}\sqrt{\left(\frac{\left(9\times {10}^{10}\right)\times \left(4.9\times {10}^{-4}\right)}{1\times \left(9\times {10}^3\right)}\right)}=35 \mathrm{Hz}$