1.

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`

Answer» Correct Answer - 11 Hz
For wire, `DeltaL=L alpha theta = (FL)/(AY)rArr D=AY alpha theta`
`:. f=(1)/(2L)sqrt((F)/(mu))=(1)/(2xx1)sqrt((10^(-6)xx2xx10^(11)xx1.21 xx10^(5)xx20)/(0.1))`
`=11Hz`


Discussion

No Comment Found

Related InterviewSolutions