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Step-by-Step Solution
Step 1: Understand the Situation
A uniform string of length $20\,\text{m}$ hangs from a rigid support. A short wave pulse is generated at the lower end of the string and travels upward. Because the string is uniform, the tension at any point depends on how much of the string is hanging below that point. As the pulse moves up, the tension it experiences changes.
Step 2: Express Tension as a Function of Position
Let us measure the distance $x$ from the rigid support at the top. Then, at a point that is $x$ meters below the support, the length of the string hanging beneath that point is $x$. If the linear mass density of the string is $\mu$, the mass of the portion of the string below that point is $\mu\,x$. Hence, the tension at that point is:
$$T = \mu \, g \, x,$$
where $g$ is the acceleration due to gravity ($g = 10\,\text{m/s}^2$, as given).
Step 3: Write the Wave Speed
The speed $v$ of a wave on a string under tension $T$ is given by:
$$v = \sqrt{\frac{T}{\mu}}.$$
Substituting $T = \mu \, g \, x$, we get:
$$v = \sqrt{\frac{\mu \, g \, x}{\mu}} = \sqrt{g \, x}.$$
This shows that the speed of the wave pulse increases as it moves higher (since $x$ increases from top to bottom).
Step 4: Compute the Time Taken Using Integration
The time $dt$ needed to travel an infinitesimal distance $dx$ at speed $v$ is $$dt = \frac{dx}{v}.$$
Therefore, the total time $t$ for the pulse to travel from the bottom ($x = 0$) to the top ($x = 20$ m) is:
$$
t = \int_{0}^{L} \frac{dx}{\sqrt{g \, x}}.
$$
We factor out the constant $\frac{1}{\sqrt{g}}$ and integrate:
$$
t = \frac{1}{\sqrt{g}} \int_{0}^{L} \frac{dx}{\sqrt{x}}
= \frac{1}{\sqrt{g}} \left[ 2\,\sqrt{x} \right]_{0}^{L}
= \frac{2\,\sqrt{L}}{\sqrt{g}}.
$$
Step 5: Substitute Numerical Values
Given $L = 20\,\text{m}$ and $g = 10\,\text{m/s}^2$, we have:
$$
t = \frac{2\,\sqrt{20}}{\sqrt{10}}
= 2 \sqrt{\frac{20}{10}}
= 2 \sqrt{2}.
$$
Hence, the time taken by the wave pulse to reach the top is
$$\boxed{2\sqrt{2}\,\text{s}}.$$
