Question

A wire of density 9 $gm/c{m}^3$ is stretched between two clamps 1.00 m apart while subjected to an extension of 0.05 cm. The lowest frequency of transverse vibrations in the wire is
(Assume Youngs modulus Y = $9\times {10}^{10}N/m^2$)

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

Answer 1

The correct option is A 35 Hz

$\rho =9gm/{cm}^{3}L=1m=100cm\Delta \ell =0.05cmNow,Y=\frac{T}{A}\frac{\ell }{\Delta \ell }\frac{T}{A}=\frac{Y\Delta \ell }{\ell }\frac{T}{A}=9\times {10}^{10}\times \frac{0.05}{10000}=45\times {10}^6\mu =m/L=\frac{\rho v}{L}=\rho \times \frac{AL}{L}=\rho A\rho =\frac{9\times {10}^6}{1000}$

Lowest frequency of transverse wave

$v=\frac{1}{2L}\sqrt{\frac{T}{\mu }}v=\frac{1}{2L}\sqrt{\frac{T}{A\rho }}v=\frac{1}{2\times 1}\sqrt{\frac{45\times {10}^6}{9\times {10}^6}\times 1000}v=\frac{1}{2}\sqrt{5000}=\frac{70}{2}=35Hz$

Hence option (A) is correct