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

For a wire under tension, the fundamental frequency is:

$$f = \frac{1}{2L}\sqrt{\frac{T}{\mu}}$$

The tension can be found from Young's modulus and strain:

$$\text{Strain} = \frac{\Delta L}{L} = \frac{3.2 \times 10^{-4}}{0.5} = 6.4 \times 10^{-4}$$

$$\text{Stress} = Y \times \text{Strain} = 8 \times 10^{10} \times 6.4 \times 10^{-4} = 5.12 \times 10^7 \text{ N/m}^2$$

The wave velocity: $$v = \sqrt{\frac{T}{\mu}} = \sqrt{\frac{\text{Stress}}{\rho}}$$ (since $$T = \text{Stress} \times A$$ and $$\mu = \rho \times A$$)

$$v = \sqrt{\frac{5.12 \times 10^7}{8 \times 10^3}} = \sqrt{6400} = 80 \text{ m/s}$$

$$f = \frac{v}{2L} = \frac{80}{2 \times 0.5} = 80 \text{ Hz}$$