A uniform rope of length 12 mm 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 wavelength 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`