Question

A steel wire of length `1m`, mass `0.1kg` and uniform cross-sectional area `10^(-6)m^(2)` is rigidly fixed at both ends. The temperature of the wire is lowered by `20^(@)C`. If transverse waves are set up by plucking the string in the middle.Calculate the frequency of the fundamental mode of vibration.
Given for steel `Y = 2 xx 10^(11)N//m^(2)`
`alpha = 1.21 xx 10^(-5) per ^(@)C`
A. 18 Hz
B. 22 Hz
C. 32 Hz
D. 42 Hz

Answer 1

Correct Answer - B

The mechanical strain
`=(Delta l)/(l)=alpha Delta T=1.21xx10^-5xx20=2.42xx10^-5`
the tension in wire
`T=Y(Delta l)/(l)A=2xx10^(11)xx2.42xx10^-5xx10^-6`
`=48.4N`
`:.` Speed of wave in wire
`V=sqrt((T)/(mu))=sqrt((48.8)/(0.1))=22m//s`
Since the wire is plucked at `(l)/(4)` from one end The wire shall oscillate in `1^(st)` overtone (for minimum number of loops)
`lamda=1m`
Now `V=f lamda` or `f=(V)/(lamda)=22Hz`.