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

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(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 mass per unit length.S

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 mass per unit length.Since the frequency $f$ of the pulse remains constant,the wavelength $\lambda$ is proportional to the wave speed $V$ $(\lambda = V/f)$.At the lower end of the rope,the tension $T_1$ is due to the mass $m_2$ only: $T_1 = m_2 g$.At the top of the rope,the tension $T_2$ is due to the combined mass of the rope and the block: $T_2 = (m_1 + m_2) g$.The ratio of wavelengths is $\frac{\lambda_2}{\lambda_1} = \frac{V_2}{V_1} = \sqrt{\frac{T_2}{T_1}}$.Substituting the values,we get $\frac{\lambda_2}{\lambda_1} = \sqrt{\frac{(m_1 + m_2) g}{m_2 g}} = \sqrt{\frac{m_1 + m_2}{m_2}}$.

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