A uniform rope of length L and mass m 1 , hangs vertically from a rigid support. A block of mass m 2 is attached to the free end of the rope. A transverse pulse of wavelength λ 1 is produced at the lower end of the rope. The wavelength of the pulse when it reaches the top of the rope is λ 2 . The ratio λ 2 λ 1 is:

Step-by-step solution

The situation in the rope is given below, The velocity of the wave at the bottom of the rope is given by, v = T μ ; where T is tension and μ is linear mass density of the string. v 1 = m 2 g L m 1 Hence, the wavelength for frequency f will be, λ 1 = v 1 f = 1 f m 2 g L m 2 Similarly, at top the velocity will be, v 1 = ( m 1 + m 2 ) g L m 1 And, the wavelength will be, λ 2 = 1 f ( m 1 + m 2 ) g L m 1 Hence, the ratio of both wavelengths will be, λ 2 λ 1 = m 1 + m 2 m 2 .