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`

Options

A. 7 Hz
B. 11 Hz
C. 15 Hz
D. 22 Hz

Answer

B

Explanation

Thermal stress in the wire ` = Y alpha Delta theta`
` :. ` Tension in the wire ` Y A alpha Delta theta " " [:' T = " stress " xx " area "] `
` :. T = 2 xx 10^(11) xx 10^(-6) xx 1.21 xx 10^(-5) xx 20 = 48.4 ` N
The mass per unit length of the wire (m)
` = (0.1)/1 = 0.1 ` kg/m
` :. ` Speed of the transverse wave in the wire,
`v = sqrt(T/m) = sqrt((48.4)/(0.1)) = 22` m/s

Since the wire is plucked at the centre,
` :. lambda =2 L `
` :. ` The fundamental frequency ,
`n = 1/(2L) sqrt(T/m) = 1/2 xx 22 = 11` Hz