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?

Answer

`(3)/(10sqrt(5))m`.

Explanation

At the bottom end
`V = sqrt((mg)/(mu)) = flambda …..(1)`
Now at the top
`V_(1) = sqrt((mg + mulg)/(mu)) = flambda' ……(2)`
`mul = M =` mass of string
From equation `(1)` & `(2)`
`lambda' = sqrt((m + M)/(m)) lambda = sqrt((2 + 8)/(2)) lambda = sqrt(5) lambda = (3)/(10sqrt(5))m`