A uniform rope of length L and mass m_1 hangs | vedclass

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Question

(B) The speed of a transverse wave on a string is given by $v = \sqrt{\frac{T}{\mu}}$,where $T$ is the tension and $\mu$ is the linear mass density.Si

Vedclass · Accepted Answer

$\sqrt{\frac{m_1 + m_2}{m_2}}$

Vedclass · Answer

(B) The speed of a transverse wave on a string is given by $v = \sqrt{\frac{T}{\mu}}$,where $T$ is the tension and $\mu$ is the linear mass density.Since the frequency $f$ of the pulse remains constant as it travels,the wavelength $\lambda = v/f$ is directly proportional to the wave speed $v$.Therefore,$\frac{\lambda_2}{\lambda_1} = \frac{v_2}{v_1} = \sqrt{\frac{T_2}{T_1}}$.At the lower end (bottom),the tension $T_1$ is due to the block of mass $m_2$ only: $T_1 = m_2 g$.At the upper end (top),the tension $T_2$ is due to the weight of both the rope and the block: $T_2 = (m_1 + m_2) g$.Substituting these values into the ratio:$\frac{\lambda_2}{\lambda_1} = \sqrt{\frac{(m_1 + m_2)g}{m_2 g}} = \sqrt{\frac{m_1 + m_2}{m_2}}$.

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