Question

A steel wire of length 1 m and mass 0.1 kg and having a uniform cross-sectional area of ${10}^{-6}{m}^2$ is rigidly fixed at both ends. The temperature of the wire is lowered by ${20}^{\circ }C$. If the transverse waves are set up by plucking the string in the middle the frequency of the fundamental note of vibration is $(Y_{steel}=2\times {10}^{11}N/m^2,{\alpha }_{steel}=1.21\times {10}^{-5}{/}^{\circ }C)$

• A. 44 Hz
• B. 88 Hz
• C. 22 Hz
• D. 11 Hz

Answer 1

The correct option is D 11 Hz

m=$0.1$kg

A=${10}^{-6}{m}^2$

$\Delta \theta ={20}^{o}C$

$Y=2\times {10}^{11}N\alpha =1.21\times {10}^{-3}{/}^{o}CNow,F=YA\alpha \Delta \theta \&\mu =\frac{\rho V}{l}=\frac{Al\rho }{l}=A\rho$

Frequency of the fundamental note of vibration is

$f=\frac{1}{\lambda }\sqrt{\frac{T}{\mu }}f=\frac{1}{2L}\sqrt{\frac{YA\alpha \Delta \theta }{A\rho }}f=\frac{1}{2\times 1}\sqrt{\frac{2\times {10}^{11}\times 1.21\times {10}^{-5}\times 20}{0.1/1}}f=\frac{1}{2}\sqrt{484\times {10}^6}f=\frac{22}{2}\times {10}^{3}f=11\times {10}^3{H}_{3}f=11{kH}_3$

Hence option (D) is correct