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A uniform rope of mass 6 kg hangs vertically from a rigid support. A block of mass 2 kg is attached to the free (lower) end of the rope. A transverse pulse of wavelength 0.06 m is generated at the lower end of the rope. What is the wavelength of the pulse when it reaches the top of the rope?
- 0.012 m
- 0.06 m
- 0.24 m
- 0.12 m
Correct answer: 0.12 m
Solution
The tension varies along the rope. At the bottom (where pulse is created), tension equals the weight of the hanging block: T_bottom = 2g = 20 N. At the top, tension equals the weight of rope + block: T_top = (6+2)g = 80 N. The wave speed v = sqrt(T/mu) where mu is linear mass density. Frequency f remains constant as the pulse travels. lambda = v/f, so lambda_top/lambda_bottom = v_top/v_bottom = sqrt(T_top/T_bottom) = sqrt(80/20) = sqrt(4) = 2. lambda_top = 2 * 0.06 = 0.12 m.
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