Question

A steel wire is rigidly fixed at both ends. Its length, mass and cross-sectional area are $1\text{m, 0.1 kg and}{10}^{-6}{\text{m}}^2$respectively. Tension in the wire is produced by lowering the temperature by ${20}^{\circ }\text{C}$. If transverse waves are set up by plucking the wire at $0.25\text{m}$ from one end, and assuming that the wire vibrates with minimum number of loops possible for such a case. The frequency of vibration (in $Hz$) is found to be $11\times \text{K}$. Find the value of $K$.$(\text{Given}\alpha =1.21\times {10}^{-5\circ }{C}^{-1},Y=2\times {10}^{11}N/m^2)$

• A. $2$
• B. $4$
• C. $11$
• D. $22$

Answer 1

The correct option is A $2$

As the temperature decreases length of wire tends to decrease by
$\Delta L=\alpha L\Delta T$
But since both ends of the wire are fixed it can't decrease its length, so stress is developed in the wire,
strain $=\frac{\Delta L}{L}=\alpha \Delta T$
Stress $=Y\cdot \frac{\Delta L}{L}=\frac{T}{a}=Y\propto \Delta T$
$\begin{array}{cc}T& =a\cdot y\alpha \Delta T\\ T& ={10}^{-6}\times 2\times {10}^{11}\times 1.21\times {10}^{-5}\times 20\end{array}=48.4N$


from the figure
$\frac{\lambda }{4}=0.25m⟹\lambda =1m$
So, standing wave is in second harmonic.

For frequncy $f=\frac{v}{\lambda }=\sqrt{\frac{T}{M/L}}\cdot \frac{1}{\lambda }$
$f=\sqrt{\frac{48.4}{0.1/1}}\times \frac{1}{1}=22$
so $k=2$