A uniform rope of length L and mass m1 hangs vertically from a rigid support. A block of mass m2 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 (a) √((m₂)/(m₁)) (b) √((m₁+m₂)/(m₁)) (c) √((m₁)/(m₂)) (d) √((m₁+m₂)/(m₂))

Answer: (d). Frequency stays the same, so λ ∝ v ∝ √(T). Bottom: T₁ = m₂ g; top: T₂ = (m₁+m₂)g. (λ₂)/(λ₁) = √((m₁+m₂)/(m₂)) So the answer is (d).

A uniform rope of length L and mass m1 hangs vertically from a rigid… | NEET 2016 (Phase 1)

Q35NEET 2016 · PHASE 1Medium

Speed of Waves

A uniform rope of length L and mass m1 hangs vertically from a rigid support. A block of mass m2 is attached to the free end of the rope. A transverse pulse of wavelength $\lambda$ 1 is produced at the lower end of the rope. The wavelength of the pulse when it reaches the top of the rope is $\lambda$ 2. The ratio $\lambda$ 2/ $\lambda$ 1 is

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Answer: (d)

Frequency stays the same, so $\lambda \propto v \propto \sqrt{T}$ .
Bottom: $T_1 = m_2 g$ ; top: $T_2 = (m_1+m_2)g$ .
$\dfrac{\lambda_2}{\lambda_1} = \sqrt{\dfrac{m_1+m_2}{m_2}}$
So the answer is (d).

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