A uniform rope of length 12 m and mass 6 kg | vedclass

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Question

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

Vedclass · Accepted Answer

$0.12$

Vedclass · Answer

(A) The speed of a transverse wave in a string is given by $v = f \lambda = \sqrt{\frac{T}{\mu}}$,where $T$ is the tension and $\mu$ is the linear mass density.Since the frequency $f$ of the pulse remains constant as it travels,we have $\lambda \propto \sqrt{T}$.Therefore,the ratio of wavelengths is $\frac{\lambda_2}{\lambda_1} = \sqrt{\frac{T_2}{T_1}}$.At the lower end (bottom),the tension $T_1$ is due to the block of mass $2 \ kg$: $T_1 = 2g$.At the top of the rope,the tension $T_2$ supports both the block $(2 \ kg)$ and the rope $(6 \ kg)$: $T_2 = (2 + 6)g = 8g$.Given $\lambda_1 = 0.06 \ m$,we calculate $\lambda_2$ as:$\lambda_2 = \lambda_1 \sqrt{\frac{T_2}{T_1}} = 0.06 	imes \sqrt{\frac{8g}{2g}} = 0.06 	imes \sqrt{4} = 0.06 	imes 2 = 0.12 \ m$.

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