A uniform rope of mass 6,kg hangs vertically | vedclass

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(B) The speed of a transverse wave in a string is given by $v = \sqrt{T/\mu} = f\lambda$,where $T$ is the tension,$\mu$ is the linear mass density,and

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$0.12$

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(B) The speed of a transverse wave in a string is given by $v = \sqrt{T/\mu} = f\lambda$,where $T$ is the tension,$\mu$ is the linear mass density,and $f$ is the frequency.Since the frequency $f$ remains constant as the pulse travels,we have $\lambda \propto \sqrt{T}$.At the bottom of the rope,the tension $T_1$ is due to the block of mass $2\,kg$,so $T_1 = 2g$.At the top of the rope,the tension $T_2$ is due to the combined mass of the rope $(6\,kg)$ and the block $(2\,kg)$,so $T_2 = (6 + 2)g = 8g$.Using the proportionality $\lambda \propto \sqrt{T}$,we get $\frac{\lambda_2}{\lambda_1} = \sqrt{\frac{T_2}{T_1}}$.Substituting the values,$\frac{\lambda_2}{0.06} = \sqrt{\frac{8g}{2g}} = \sqrt{4} = 2$.Therefore,$\lambda_2 = 0.06 	imes 2 = 0.12\,m$.

$A$ uniform rope of mass $6\,kg$ hangs vertically from a rigid support. $A$ block of mass $2\,kg$ is attached to the free end of the rope. $A$ transverse pulse of wavelength $0.06\,m$ is produced at the lower end of the rope. The wavelength of the pulse when it reaches the top is (in $m$):

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