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

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(A) Let the velocity of the pulse at the lower end be $v_1$ and at the top be $v_2$. Since the frequency $n$ of the wave remains constant,we have $\la

Vedclass · Accepted Answer

$\left[\frac{m_2}{m_1+m_2}\right]^{\frac{1}{2}}$

Vedclass · Answer

(A) Let the velocity of the pulse at the lower end be $v_1$ and at the top be $v_2$. Since the frequency $n$ of the wave remains constant,we have $\lambda = \frac{v}{n}$,which implies $\frac{\lambda_2}{\lambda_1} = \frac{v_2}{v_1}$. The velocity 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. Thus,$v \propto \sqrt{T}$. At the lower end,the tension $T_1$ is due to the block of mass $m_2$,so $T_1 = m_2 g$. At the top end,the tension $T_2$ is due to the total mass of the rope and the block,so $T_2 = (m_1 + m_2) g$. Therefore,$\frac{\lambda_2}{\lambda_1} = \frac{v_2}{v_1} = \sqrt{\frac{T_2}{T_1}} = \sqrt{\frac{(m_1 + m_2)g}{m_2 g}} = \sqrt{\frac{m_1 + m_2}{m_2}}$. Taking the reciprocal,we get $\frac{\lambda_1}{\lambda_2} = \sqrt{\frac{m_2}{m_1 + m_2}} = \left[\frac{m_2}{m_1+m_2}\right]^{\frac{1}{2}}$.

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