Question

A steel wire is rigidly fixed at both ends. Its length, mass and cross-sectional area are $1m,0.1kg$ and ${10}^{-6}{m}^2$ respectively. Then the temperature of the wire is lowered by $20{}^{o}C$. If the transverse waves are setup by plucking the wire at $0.25m$ from one end and assuming that wire vibrates with minimum number of loops possible for such a case. Find the frequency of vibration.
[coefficient of linear expansion of steel $=1.21\times {10}^{-5}{/}^{o}C$ and Young’s modulus $=2\times {10}^{11}N/m^2$]

• A. $32Hz$
• B. $22Hz$
• C. $18Hz$
• D. $42Hz$

Answer 1

The correct option is B $22Hz$

The mechanical strain, $\frac{\Delta l}{l}=a\Delta T$
$=1.21\times {10}^{-5}\times 20=2.42\times {10}^{-4}$
The tension in wire
$\Rightarrow T=Y\frac{\Delta l}{l}A$
$T=2\times {10}^{11}\times 2.42\times {10}^{-4}\times {10}^{-4}$
$=48.4N$
$\therefore$ Speed of wave in wire
$V=\sqrt{\frac{T}{\mu }}=\sqrt{\frac{48.4}{0.1}}=22m/s$
Since the wire is plucked at $\frac{l}{4}$ from one end


The wire shall oscillate in $1st$ overtone (for minimum number of loops)
$\lambda =l=1m$
Now $V=f\lambda$
or $f=\frac{V}{\lambda }=22Hz$

Final answer: (b)