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Chapter 16: Problem 75

(a) What is the fastest transverse wave that can be sent along a steel wire? For safety reasons, the maximum tensile stress to which steel wires should be subjected is \(7.00 \times 10^{8} \mathrm{~N} / \mathrm{m}^{2}\). The density of steel is \(7800 \mathrm{~kg} / \mathrm{m}^{3} .\) (b) Does your answer depend on the diameter of the wire?

Short Answer

Expert verified
The wave speed is approximately 299.57 m/s and it does not depend on the wire's diameter.

Step by step solution

01

Understand the Concept

The speed of a transverse wave in a stretched wire can be found using the formula \( v = \sqrt{\frac{T}{\mu}} \), where \( T \) is the tension in the wire and \( \mu \) is the mass per unit length of the wire. The tension \( T \) is related to the stress and cross-sectional area.
02

Calculate the Tension

The tension \(T\) in the wire can be found using the formula for stress: \( \text{Stress} = \frac{T}{A} \), where \(A\) is the cross-sectional area. Rearranging, \( T = \text{Stress} \times A \). The maximum tensile stress given is \( 7.00 \times 10^{8} \mathrm{~N/m^2} \). Hence, \( T = 7.00 \times 10^{8} \times A \).
03

Express Mass Per Unit Length

The mass per unit length \( \mu \) is given by \( \mu = \rho \times A \), where \( \rho \) is the density \(7800 \, \text{kg/m}^3\).
04

Derive the Wave Speed Formula

Substitute the given expressions for \( T \) and \( \mu \) into the wave speed formula: \[ v = \sqrt{\frac{7.00 \times 10^{8} \times A}{7800 \times A}} \]. Simplifying this expression gives \[ v = \sqrt{\frac{7.00 \times 10^{8}}{7800}} \].
05

Compute the Wave Speed

Calculate the speed: \[ v = \sqrt{\frac{7.00 \times 10^{8}}{7800}} = \sqrt{89743.59} \]. Hence, \( v \approx 299.57 \, \text{m/s} \).
06

Check the Dependence on Diameter

Since both the tension and mass per unit length have \( A \) (cross-sectional area) as a common factor, the diameter (or area \( A \)) cancels out. Thus, the wave speed \( v \) is independent of the diameter of the wire.

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Key Concepts

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

Transverse Waves
Transverse waves can be thought of as waves that move perpendicular to the direction of the wave itself. Imagine a rope being flicked up and down; the waves travel along the length of the rope while moving side to side.
In the context of steel wires, transverse waves are the oscillations that occur due to vibrations within the wire. The speed at which these waves travel is crucial for understanding how quickly energy is transmitted through the wire.
This speed is determined by the material properties and the tension within the wire. Lower tension or heavier wires will lead to slower wave speeds.
Tensile Stress
Tensile stress refers to the force exerted on a material along its length, intending to stretch it. In steel wires, tensile stress plays a key role in determining the maximum limit of force application.
To calculate tensile stress, we use the formula:
  • Tensile Stress \( = \frac{T}{A} \)
  • Where \(T\) is the tension in the wire, and \(A\) is the cross-sectional area.
For steel wires, there is a safety threshold for tensile stress, which means it's important not to exceed a value of \(7.00 \times 10^8 \text{ N/m}^2\). Exceeding this could risk breaking or permanently deforming the wire.
Density of Steel
The density of steel is an important physical property when calculating wave speed. Density is a measure of mass per unit volume and directly influences the mass per unit length of the wire. For steel, the density is given as \(7800 \text{ kg/m}^3\).
Knowing the density, we can determine how much mass is in a given volume of steel wire, which affects how the wire responds to applied forces. Higher density means more mass within a certain volume, thus affecting the speed at which waves travel through the material.
Mass Per Unit Length
Mass per unit length, often represented as \( \mu \), is derived from the density of the material and its cross-sectional area. It provides a measure of how much mass exists in a given length of the wire.This is calculated as:
  • \( \mu = \rho \times A \)
  • Where \( \rho \) is the density of the material, and \( A \) is the cross-sectional area.
In problems involving wave speed, knowing the mass per unit length helps determine how quickly a wave can travel through the material. In essence, it plays a crucial role in the physics of vibrations and oscillations.

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Most popular questions from this chapter

93 A traveling wave on a string is described by $$ y=2.0 \sin \left[2 \pi\left(\frac{t}{0.40}+\frac{x}{80}\right)\right] $$ where \(x\) and \(y\) are in centimeters and \(t\) is in seconds. (a) For \(t=0\), plot \(y\) as a function of \(x\) for \(0 \leq x \leq 160 \mathrm{~cm} .\) (b) Repeat (a) for \(t=0.05 \mathrm{~s}\) and \(t=0.10 \mathrm{~s}\). From your graphs, determine (c) the wave speed and (d) the direction in which the wave is traveling.

In Fig. \(16-42\), a string, tied to a sinusoidal oscillator at \(P\) and running over a support at \(Q\), is stretched by a block of mass \(m\). The separation \(L\) between \(P\) and \(Q\) is \(1.20 \mathrm{~m}\), and the frequency \(f\) of the oscillator is fixed at \(120 \mathrm{~Hz}\). The amplitude of the motion at \(P\) is small enough for that point to be considered a node. A node also exists at \(Q .\) A standing wave appears when the mass of the hanging block is \(286.1 \mathrm{~g}\) or \(447.0 \mathrm{~g} .\) but not for any intermediate mass. What is the linear density of the string?

A human wave. During sporting events within large, densely packed stadiums, spectators will send a wave (or pulse) around the stadium (Fig. \(16-29)\). As the wave reaches a group of spectators, they stand with a cheer and then sit. At any instant, the width \(w\) of the wave is the distance from the leading edge (people are just about to stand) to the trailing edge (people have just sat down). Suppose a human wave travels a distance of 853 seats around a stadium in \(39 \mathrm{~s}\), with spectators requiring about \(1.8 \mathrm{~s}\) to respond to the wave's passage by standing and then sitting. What are (a) the wave speed \(v\) (in seats per second) and (b) width \(w\) (in number of seats)?

The linear density of a string is \(1.6 \times 10^{-4} \mathrm{~kg} / \mathrm{m} .\) A transverse wave on the string is described by the equation $$ y=(0.021 \mathrm{~m}) \sin \left[\left(2.0 \mathrm{~m}^{-1}\right) x+\left(30 \mathrm{~s}^{-1}\right) t\right] $$ What are (a) the wave speed and (b) the tension in the string?

Two sinusoidal waves of the same period, with amplitudes of \(5.0\) and \(7.0 \mathrm{~mm}\), travel in the same direction along a stretched string; they produce a resultant wave with an amplitude of \(9.0 \mathrm{~mm}\). The phase constant of the \(5.0 \mathrm{~mm}\) wave is \(0 .\) What is the phase constant of the \(7.0 \mathrm{~mm}\) wave?
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