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Topic: Wave motion: longitudinal and transverse waves, wave speed· Oscillations and Waves

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 whe…

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 \(\lambda_1\) 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 :

1. \(\sqrt{\frac{m_1}{m_2}}\)

2. \(\sqrt{\frac{m_1+m_2}{m_2}}\)

3. \(\sqrt{\frac{m_2}{m_1}}\)

4. \(\sqrt{\frac{m_1+m_2}{m_1}}\)

1 answer

Best answer

Answer: 2. \(\sqrt{\frac{m_1+m_2}{m_2}}\)

Key idea: the frequency of a pulse stays the same, so wavelength is proportional to wave speed \(v=\sqrt{T/\mu}\).

  1. At the lower end, tension \(T_1=m_2g\).
  2. At the top, tension \(T_2=(m_1+m_2)g\).
  3. \(\dfrac{\lambda_2}{\lambda_1}=\dfrac{v_2}{v_1}=\sqrt{\dfrac{T_2}{T_1}}=\sqrt{\dfrac{m_1+m_2}{m_2}}\).
Doubtly AI · 30 Oct 2026 AI-generated · not yet verified Report
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