A uniform wire of length 12 m and mass 6 kg is suspended from a rigid support. A mass of 2 kg is a attached to the lower free end. A transverse wave of length 0 . 0 6 m is generated at the lower end of the wire . What is its wavelength when the wave reaches the upper end ?

Explanation

Velocity of transverse vibration in a stretched wire, ` V= sqrt((T)/(m))` , where T = tension in the wire and m = mass per unit length = constant, for the uniform wire.
If `V_(1) and V_(2)` are the velocities at the lower and upper ends, respectively,
`(V_(1))/(V_(2)) = sqrt((T_(1))/(T_(2)))`
Here, `T_(1)=` tension at the lower end = weight of the suspended mass ` 2 xx 9. 8 ` N ,
`T_(2)=` tension at the upper end = weight of the wire and the suspended mass ` = ( 2 + 6) xx 9 . 8 = 8 xx 9 . 8 ` N
The entire wire will emit a sound of a fixed frequency, so the wavelength `lambda` is proportional to the wave velocity V .
`:. lambda_(2) = (V_(1))/(V_(2)) = sqrt((T_(1))/(T_(2)))`
or, ` lambda_(2) = lambda_(1) sqrt((T_(2))/(T_(1))) = 0 . 06 xx sqrt((8 xx 9 . 8)/(2 xx 9 . 8)) = 0 . 12 ` m .