# 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 wave length 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?

10 years ago
Hello Student,
KEY CONCEPT: The velocity of wave on the string is given by the formula
v = √T/m
Where t is the tension and m is the mass per unit length. Since the tension in the string will increase as we move up the string (as the string has mass), therefore the velocity of wave will also increase. (m is the same as the rope is uniform )
∴ v1/v2 = √T1/T2 = √2 x 9.8/8 x 9.8 = 1/2 ∴ v2 = 2v1
Since frequency remains the same
∴ ℷ2 = 2ℷ1 = 2 x 0.06 = 0.12 m

Thanks
SHAIK AASIF AHAMED
10 years ago
Hello student,
vtop/vbottom=sqrt(ttop/tbottom)=sqrt(6+2/2)=2
So f$\lambda$top/f$\lambda$bottom=2
as frequency is same we get$\lambda$top=2($\lambda$bottom)=2(0.06)=0.12
Thanks and Regards
Shaik Aasif
Yash Chourasiya
4 years ago
Dear Student

As rope is heavy, Tension is different at different points.Tat lower end is2gand at upper end is6g
V=vλ