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Chapter 11: Problem 11

A uniform string of length \(10.0 \mathrm{m}\) and weight \(0.25 \mathrm{N}\) is attached to the ceiling. A weight of \(1.00 \mathrm{kN}\) hangs from its lower end. The lower end of the string is suddenly displaced horizontally. How long does it take the resulting wave pulse to travel to the upper end? [Hint: Is the weight of the string negligible in comparison with that of the hanging mass?]

Short Answer

Expert verified
The time it takes for the wave pulse to travel from the lower end to the upper end of the string is approximately 0.05 seconds.

Step by step solution

01

Determine the tension on the string

First, we need to calculate the tension on the string, which is caused by the weight of the hanging mass. Since the weight of the mass is \(1.00 \mathrm{kN}\), its force in newtons is \(F_{mass} = 1.00 * 1000 \mathrm{N}\). Since the string is in equilibrium, the tension in the string is equal to the sum of the force due to the hanging mass and the force due to the weight of the string itself: \(T = F_{mass} + F_{string} = 1000\mathrm{N}+0.25\mathrm{N} = 1000.25\mathrm{N}\).
02

Calculate the wave speed

Next, we calculate the wave speed using the formula: \(v = \sqrt{\frac{T}{\mu}}\), where \(v\) is the wave speed, \(T\) is the tension, and \(\mu\) is the linear mass density of the string. We know the tension and the total weight of the string from the problem (\(0.25 \mathrm{N}\)), so we can calculate the linear mass density by dividing the total weight by the length of the string: \(\mu = \frac{0.25\mathrm{N}}{10\mathrm{m}} = 0.025 \mathrm{N/m}\). Now we can calculate the wave speed: \(v = \sqrt{\frac{1000.25\mathrm{N}}{0.025 \mathrm{N/m}}} = \sqrt{40010} \mathrm{m/s} \approx 200.03 \mathrm{m/s}\).
03

Calculate the time it takes for the wave pulse to travel up the string

Finally, we can find the time it takes for the wave pulse to travel up the string using the formula \(t = \frac{d}{v}\), where \(t\) is the time, \(d\) is the distance (which is the length of the string, \(10\mathrm{m}\)), and \(v\) is the wave speed we calculated in step 2. Thus, \(t = \frac{10\mathrm{m}}{200.03\mathrm{m/s}} \approx 0.05\mathrm{s}\). So, the time it takes for the wave pulse to travel from the lower end to the upper end of the string is approximately \(0.05 \mathrm{s}\).

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