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 ${20}^{0}C.$ 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.
[Coefficient of linear expansion of steel $=1.21\times {10}^{-5/0}C$ and Young's modulus $=2\times {10}^{11}N/m^2$]

• A. The tension in the wire will be 48.4 N.
• B. The number of loops will be 2.
• C. The frequency of vibration will be 22 Hz.
• D. The frequency of vibration will be 44 Hz.

Answer 1

Answer: (A), (B), (C)

The correct options are
A The tension in the wire will be 48.4 N.
B The number of loops will be 2.
C The frequency of vibration will be 22 Hz.

The mechanical strain
$=\frac{\Delta l}{l}=\alpha \Delta T=1.21\times {10}^{-5}\times 20=2.42\times {10}^{-5}$
The tension in wire
$=T=Y\frac{\Delta l}{l}A=2\times {10}^{11}\times 2.42\times {10}^{-5}\times {10}^{-6}=48.4N$
$\therefore$ speed of wave 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 $1^{st}$ overtone (for minimum number of loops)
$\lambda =l=1m$
Now $V=f\lambda$ or $f=\frac{V}{\lambda }=22$ Hz.