Question

A wire of density 9$gm/c{m}^2$ is stretched between two clamps 1.00 m apart subjected extension of 0.05 cm. The lowest frequency of transverse vibrations in the wire is
(Assume Young's modulus Y = $9\ast {10}^{1}0N/m^2$)

• A. 35 Hz
• B. 45 Hz
• C. 75 Hz
• D. 90 Hz

Answer 1

The correct option is A 35 Hz

$\begin{array}{c}Themassperunitlength,\\ m=\frac{M}{l}=\frac{Al\rho }{l}=A\rho \\ ulesofelasticity\\ y=\frac{\frac{T}{A}}{\frac{\Delta l}{l}}\\ T=\frac{Y\Delta lA}{l}\\ Hencelowestfrequencyofvibration\\ n=\frac{1}{2l}\sqrt{\frac{T}{m}}\\ =\frac{1}{2l}\sqrt{\frac{y\left(\frac{\Delta l}{l}\right)A}{A\rho }}=\frac{1}{2l}\sqrt{\frac{y\Delta l}{l\rho }}\\ n=\frac{1}{2\times 1}\sqrt{\frac{9\times {10}^{10}\times 4.9\times {10}^{-4}}{1\times 9\times {10}^3}}\\ =35Hz\end{array}$