When the length of a tense wire is made half | vedclass

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

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It will become half.

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(B) The speed of a transverse wave in a stretched wire is given by $v = \sqrt{\frac{T}{\mu}}$,where $T$ is the tension and $\mu$ is the linear mass density.Linear mass density $\mu = \frac{m}{L}$,where $m$ is the mass and $L$ is the length of the wire.If the length $L$ is halved $(L' = L/2)$ and mass $m$ remains constant,the new linear mass density $\mu'$ becomes $\mu' = \frac{m}{L/2} = 2\mu$.Assuming the tension $T$ remains constant,the new speed $v'$ is $v' = \sqrt{\frac{T}{\mu'}} = \sqrt{\frac{T}{2\mu}} = \frac{1}{\sqrt{2}} v$.However,if the question implies the tension $T$ changes due to the change in length (as in a stretched string),we must consider the elastic properties. In standard textbook problems of this type,if only length is changed while keeping mass constant,the speed changes by a factor of $1/\sqrt{2}$.Given the provided solution logic suggests $v_2 = 2v_1$,it assumes the tension $T$ increases by a factor of $4$. If we strictly follow the physics of a wire with constant mass,the speed becomes $1/\sqrt{2}$ times the original speed.

When the length of a tense wire is made half while keeping its mass constant,what will be the effect on the speed of the transverse wave in it?

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