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

Vedclass · Accepted Answer

$v \sqrt{2}$

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(C) The speed of a transverse wave in a stretched string is given by $v = \sqrt{\frac{T}{\mu}}$,where $T$ is the tension and $\mu$ is the linear mass density.According to Hooke's Law,the tension $T$ in a stretched string is proportional to the extension $\Delta l$,so $T = k \Delta l$.Since the mass and length of the string remain constant,$\mu$ is constant. Thus,$v \propto \sqrt{T} \propto \sqrt{\Delta l}$.Given the initial extension $\Delta l_1 = \frac{L}{20}$ and initial speed $v_1 = v$.For the second case,the extension is $\Delta l_2 = \frac{L}{10}$.Using the ratio: $\frac{v_2}{v_1} = \sqrt{\frac{\Delta l_2}{\Delta l_1}} = \sqrt{\frac{L/10}{L/20}} = \sqrt{\frac{20}{10}} = \sqrt{2}$.Therefore,$v_2 = v \sqrt{2}$.

$A$ string of length $L$ is stretched by $\frac{L}{20}$ and the speed of transverse waves along it is $v$. The speed of the wave when it is stretched by $\frac{L}{10}$ will be (assume that Hooke's law is applicable).

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