A uniform rope of length `12 m` and mass ` 6 kg` hangs vertically from a rigid support . A block of mass `2 kg` is attached to the free end of the rope . A transverse pulse of wavelengths `0.06 m` is produced at the lower end of the rope . What is the wavelength of the pulse when it reaches the top of the rope?

Explanation

Tension at the lower end of the rope ,
`T_(1) = 2 g = 2 xx 9.8 = 19.6 N`
Tension at the upper end of rope ,
Let `v_(1) and v_(2)` be the speeds of pulse at the lower and upper end , respectively . So
`v = sqrt (((T_(1))/(m))), v_(2) = sqrt(((T_(2))/(m)))`
On dividing , we get
`(v_(2))/( v_(1)) = sqrt(((T_(2))/( T_(1)))) = sqrt(((78.4)/(19.6))) = sqrt(4) = 2`
As frequency is independent of medium , therefore if `lambda_(1) and lambda_(2)` are wavelengths at lower and upper ends respectively . Then
`v_(1) = n lambda_(1) and v_(2) = n lambda_(2)`
So, `(lambda_(2))/(lambda_(1)) = (v_(2))/( v_(1)) = 2`
Therefore , the wavelength of pulse at upper end ` = 2 lambda`.
`= 2 xx 0.06 = 0.12 m`