Question

A wire of density $9\times {10}^{3}kg/m^3$ is stretched between two clamps 1 m apart and is subjected to an extension of $4.9\times {10}^{-4}m$. The lowest frequency of transverse vibration in the wire is $(Y=9\times {10}^{10}N/m^2)$

• A. 40 Hz
• B. 35 Hz
• C. 30 Hz
• D. 25 Hz

Answer 1

The correct option is B

35 Hz

Then mass per unit length Then mass per unit length $m=\frac{M}{l}=\frac{Al\rho }{l}=A\rho$

And Young’s modules of elasticity $Y=\frac{\frac{T}{A}}{\Delta \frac{l}{l}}$

$\Rightarrow T=\frac{Y\Delta lA}{l}$. Hence lowest frequency of vibration $n=\frac{1}{2l}\sqrt{\frac{T}{m}}=\frac{1}{2l}\sqrt{\frac{Y\left(\frac{\Delta l}{l}\right)A}{A\rho }}=\frac{1}{2l}\sqrt{\frac{Y\Delta l}{l\rho }}$

$\Rightarrow n=\frac{1}{2\times 1}\sqrt{\frac{9\times {10}^{10}\times 4.9\times {10}^{-4}}{1\times 9\times {10}^3}}=35Hz$