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(D) The speed 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.At

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both $(B)$ and $(C)$

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(D) The speed 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.At a distance $x$ from the lower end,the tension $T(x) = w + \mu x g$,where $w$ is the weight attached at the lower end.Thus,$v = \sqrt{\frac{w + \mu x g}{\mu}} = \sqrt{\frac{w}{\mu} + xg}$.As the pulse travels upwards,$x$ increases,so the speed $v$ increases.Squaring both sides,$v^2 = \frac{w}{\mu} + xg$.Differentiating with respect to time $t$,$2v \frac{dv}{dt} = g \frac{dx}{dt}$.Since $\frac{dx}{dt} = v$,we have $2v \frac{dv}{dt} = gv$,which simplifies to $\frac{dv}{dt} = \frac{g}{2}$.Thus,the pulse travels upwards with a constant acceleration of $\frac{g}{2}$.

$A$ perfectly elastic uniform string is suspended vertically with its upper end fixed to the ceiling and the lower end loaded with a weight. If a transverse wave is imparted to the lower end of the string,the pulse will

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