Question

A wire of density 9x10^3 kg/m^3 is stretched between two clamps 1m apart and is subjected to an extension of 4.9x10^-4m .What will be the lowest frequency of the vibrations?

Answer 1

$$\begin{gathered} Weknowthelowestfrequencyproducedinbetweenthetwowedgesisgivenby; \\ \nu =\frac{1}{2L}\sqrt{\frac{T}{m}}=\frac{1}{2L}\sqrt{\frac{Y\times \pi r^2\times ∆l/L}{{\mathrm{\pi r}}^2\mathrm{L}\times \mathrm{\rho }}}=\frac{1}{2L}\sqrt{\frac{Y\times ∆l}{L^2\times \rho }} \\ WeknowL=1m,Y=2\times {10}^{11}N/m^2,\rho =9\times {10}^{3}Kg/m^{3}and∆l=4.9\times {10}^{-4}m. \\ So\nu =\frac{1}{2}\sqrt{\frac{2\times {10}^{11}\times 4.9\times {10}^{-4}}{9\times {10}^3}}=\frac{1\times 2\times 10\times 2.24\times 7}{2\times 3}=51.5HZ. \end{gathered}$$