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

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(A) 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

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

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(A) 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,the wavelength $\lambda = \frac{v}{f}$ is directly proportional to the velocity $v$. Therefore,$\lambda \propto \sqrt{T}$.At the lower end of the rope (where the block $m_2$ is attached),the tension $T_1$ is due to the weight of the block: $T_1 = m_2 g$.At the top of the rope,the tension $T_2$ is due to the weight of both the rope and the block: $T_2 = (m_1 + m_2) g$.Thus,the ratio of the wavelengths is:$\frac{\lambda_2}{\lambda_1} = \sqrt{\frac{T_2}{T_1}} = \sqrt{\frac{(m_1 + m_2) g}{m_2 g}} = \sqrt{\frac{m_1 + m_2}{m_2}}$.

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