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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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Most popular questions from this chapter

The speed of sound in air at room temperature is \(340 \mathrm{m} / \mathrm{s}\) (a) What is the frequency of a sound wave in air with wavelength $1.0 \mathrm{m} ?$ (b) What is the frequency of a radio wave with the same wavelength? (Radio waves are electromagnetic waves that travel at $3.0 \times 10^{8} \mathrm{m} / \mathrm{s}$ in air or in vacuum.)
(a) Sketch graphs of \(y\) versus \(x\) for the function $$ y(x, t)=(0.80 \mathrm{mm}) \sin (k x-\omega t) $$ for the times \(t=0,0.96 \mathrm{s},\) and \(1.92 \mathrm{s} .\) Make all three graphs of the same axes, using a solid line for the first, a dashed line for the second, and a dotted line for the third. Use the values \(k=\pi /(5.0 \mathrm{cm})\) and $\omega=(\pi / 6.0) \mathrm{rad} / \mathrm{s}$ (b) Repeat part (a) for the function $$ y(x, t)=(0.50 \mathrm{mm}) \sin (k x+\omega t) $$ (c) Which function represents a wave traveling in the \(-x\) direction and which represents a wave traveling in the \(+x\) -direction?
The equation of a wave is $$ y(x, t)=(3.5 \mathrm{cm}) \sin \left\\{\frac{\pi}{3.0 \mathrm{cm}}[x-(66 \mathrm{cm} / \mathrm{s}) t]\right\\} $$ Find (a) the amplitude and (b) the wavelength of this wave.
A transverse wave on a string is described by the equation $y(x, t)=(2.20 \mathrm{cm}) \sin [(130 \mathrm{rad} / \mathrm{s}) t+(15 \mathrm{rad} / \mathrm{m}) x].$ (a) What is the maximum transverse speed of a point on the string? (b) What is the maximum transverse acceleration of a point on the string? (c) How fast does the wave move along the string? (d) Why is your answer to (c) different from the answer to (a)?
A seismic wave is described by the equation $y(x, t)=(7.00 \mathrm{cm}) \cos [(6.00 \pi \mathrm{rad} / \mathrm{cm}) x+(20.0 \pi \mathrm{rad} / \mathrm{s}) t]\( The wave travels through a uniform medium in the \)x$ -direction. (a) Is this wave moving right ( \(+x\) -direction) or left (-x-direction)? (b) How far from their equilibrium positions do the particles in the medium move? (c) What is the frequency of this wave? (d) What is the wavelength of this wave? (e) What is the wave speed? (f) Describe the motion of a particle that is at \(y=7.00 \mathrm{cm}\) and \(x=0\) when \(t=0 .\) (g) Is this wave transverse or longitudinal?
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