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Chapter 17: Problem 11

What Is the Speed What is the speed of a transverse wave in a rope of length \(2.00 \mathrm{~m}\) and mass \(60.0 \mathrm{~g}\) under a tension of \(500 \mathrm{~N}\) ?

Short Answer

Expert verified
The speed of the transverse wave is \(129.1 \text{ m/s} \).

Step by step solution

01

- Identify the given values

First, note the given values from the problem: length of the rope ((`L`) = 2.00 m, mass of the rope (`m`) = 60.0 g = 0.060 kg, tension (`T`) = 500 N.
02

- Calculate the linear mass density

The linear mass density ((`\mu`) is calculated as \(\mu = \frac{m}{L}\). Substituting the given values: \[ \mu = \frac{0.060 \text{ kg}}{2.00 \text{ m}} = 0.030 \text{ kg/m} \]
03

- Use the wave speed formula

The speed ((`v`) of a transverse wave on a rope can be found using the formula: \( v = \sqrt{\frac{T}{\mu}} \).
04

- Substitute the values and solve

Substituting the known values ((`T = 500 N`) and ((`\mu = 0.030 \text{ kg/m}`) into the wave speed formula: \[ v = \sqrt{\frac{500 \text{ N}}{0.030 \text{ kg/m}}} = \sqrt{16666.67 \text{ m}^2/\text{s}^2} \approx 129.1 \text{ m/s} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Wave Speed Formula
To calculate the speed of a transverse wave on a rope, we use the wave speed formula, which is:
\[ v = \sqrt{\frac{T}{\mu}} \]
In this formula, \( v \) represents the speed of the wave. The tension in the rope is denoted by \( T \), and \( \mu \) is the linear mass density of the rope.

This formula shows how wave speed is related to the tension and linear mass density. The higher the tension, the faster the wave travels. Conversely, the higher the mass per unit length, the slower the wave. This relationship will be useful to check how different factors affect wave speed.
Linear Mass Density
Linear mass density, \( \mu \), indicates how much mass is distributed along a unit length of the rope. It can be found using the formula:
\[ \mu = \frac{m}{L} \]
Here, \( m \) is the mass of the rope and \( L \) is its length.

For example, in this problem, we found it as follows:
\[ \mu = \frac{0.060 \text{ kg}}{2.00 \text{ m}} = 0.030 \text{ kg/m} \]
This calculation tells us that for every meter of the rope, the mass is 0.030 kg. Understanding linear mass density can help us evaluate how the weight distribution in the rope affects the wave behavior. Greater linear mass density means more resistance to the motion of the wave.
Tension in Rope
Tension, denoted as \( T \), is the force applied along the length of the rope. In this problem, the given tension is \( 500 \text{ N} \). Tension plays a critical role in determining wave speed.

When a rope is under tension, it becomes taut, allowing the wave to travel through it more efficiently. The higher the tension, the stronger the force pulling the rope tight, which in turn increases the speed of the wave traveling through it.

Let's put this into perspective using our given problem. By substituting the tension (\( T = 500 \text{ N} \)) and linear mass density (\( \mu = 0.030 \text{ kg/m} \)) into the wave speed formula, we calculated the speed as:
\[ v = \sqrt{\frac{500 \text{ N}}{0.030 \text{ kg/m}}} = \sqrt{16666.67 \text{ m}^2/\text{s}^2} \approx 129.1 \text{ m/s} \]
This tells us that as tension increases, so does the wave speed, making the wave travel faster along the rope.

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Most popular questions from this chapter

Two Identical Traveling Waves Two identical traveling waves, moving in the same direction, are out of phase by \(\pi / 2\) rad. What is the amplitude of the resultant wave in terms of the common amplitude \(Y\) of the combining waves?

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Explaining the Wave Equation The wave equation $$ \frac{\partial^{2} y}{\partial x^{2}}=\frac{1}{v_{1}^{2}} \frac{\partial^{2} y}{\partial t^{2}} $$ is often used to describe the transverse displacement of waves on a stretched spring. Explain the meaning of each of the elements of this equation with reference to the physical spring and discuss under what circumstances you expect it to be a good description.

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