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A heavy uniform rope is suspended vertically from a ceiling and is in equilibrium. A pulse is generated at the bottom end of the rope as shown. As the pulse travels up the rope, its acceleration at any instant is
( $\mathrm{g}$ is acceleration due to gravity)
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( $\mathrm{g}$ is acceleration due to gravity)
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Verified Answer
The correct answer is:
Constant and equal to $\frac{g}{2}$
Consider a heavy rope which have mass $\mathrm{M}$ and length $\mathrm{L}$.
$\mathrm{a}=\mathrm{x} \text { (constant) }$
Tension in rope, $\mathrm{T}=\frac{\mathrm{M}}{\mathrm{L}} \mathrm{xg}=\mu \mathrm{xg}$
speed of pulse, $v=\sqrt{\frac{i}{\mu}}$
$\begin{aligned}
& =\sqrt{\frac{\mu x g}{\mu}}=\sqrt{x g} \\
& \frac{d v}{d t}=\sqrt{g} \times \frac{1}{2 \sqrt{x}} \frac{d x}{d i} \\
& =\frac{1}{2} \sqrt{\frac{g}{x}} v \\
& =\frac{1}{2} \sqrt{\frac{g}{x}} \times \sqrt{x g} \\
& \frac{d v}{d t}=\frac{g}{2}
\end{aligned}$
Acceleration, $\mathrm{a}=\frac{\mathrm{g}}{2}$
$\mathrm{a}=\mathrm{x} \text { (constant) }$
Tension in rope, $\mathrm{T}=\frac{\mathrm{M}}{\mathrm{L}} \mathrm{xg}=\mu \mathrm{xg}$
speed of pulse, $v=\sqrt{\frac{i}{\mu}}$
$\begin{aligned}
& =\sqrt{\frac{\mu x g}{\mu}}=\sqrt{x g} \\
& \frac{d v}{d t}=\sqrt{g} \times \frac{1}{2 \sqrt{x}} \frac{d x}{d i} \\
& =\frac{1}{2} \sqrt{\frac{g}{x}} v \\
& =\frac{1}{2} \sqrt{\frac{g}{x}} \times \sqrt{x g} \\
& \frac{d v}{d t}=\frac{g}{2}
\end{aligned}$
Acceleration, $\mathrm{a}=\frac{\mathrm{g}}{2}$
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