Question

A steel wire of length, mass and cross sectional area of 1 $m$, 0.1 $kg$ and ${10}^{-6}{m}^2$ respectively is rigidly fixed at both ends. Now, if the temperature of the wire is lowered by ${20}^{\circ }C$ and transverse waves are setup by plucking the wire at 0.25 $m$ from one end. Assuming that the wire vibrates with minimum number of loops possible, for such case the frequency of vibration in $Hz$ is $11x$. Then the value of $x$ is . Write upto two digits after the decimal point.
[${\alpha }_{steel}=1.21\times {10}^{-5}{/}^{\circ }C$ and $Y=2.0\times {10}^{11}N/m^2$ ]

Answer 1

We know that the mechanical strain $=\frac{\Delta l}{l}$.

$\Rightarrow \frac{\Delta l}{l}=\alpha \Delta T=1.21\times {10}^{-5}\times 20=2.42\times {10}^{-5}$

Let $T$ be the tension in wire
$\therefore T=Y\frac{\Delta l}{l}A=2\times {10}^{11}\times 2.42\times {10}^{-5}\times {10}^{-6}=48.4N$

$\therefore$ Speed of wave $V$ in wire is$V=\sqrt{\frac{T}{\mu }}=\sqrt{\frac{48.4}{0.1}}=22m/s$

Since the wire is plucked at $\frac{l}{4}$ from one end, the wire shall oscillate in $1^{st}$ overtone (for minimum number of loops)

$\therefore$ $\lambda =l=1m$
Now $V=f\lambda$
$f=\frac{V}{\lambda }=22=11\left(2\right)Hz$
$\therefore$ The value of $x$ is 2.