/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 55 The following two waves are sent... [FREE SOLUTION] | 91Ó°ÊÓ

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The following two waves are sent in opposite directions on a horizontal string so as to create a standing wave in a vertical plane: $$ \begin{array}{l} y_{1}(x, t)=(6.00 \mathrm{~mm}) \sin (4.00 \pi x-400 \pi t) \\ y_{2}(x, t)=(6.00 \mathrm{~mm}) \sin (4.00 \pi x+400 \pi t) \end{array} $$ with \(x\) in meters and \(t\) in seconds. An antinode is located at point \(A\). In the time interval that point takes to move from maximum upward displacement to maximum downward displacement, how far does each wave move along the string?

Short Answer

Expert verified
Each wave moves 0.25 m.

Step by step solution

01

Calculate the Standing Wave Equation

A standing wave can be represented by the sum of two opposite traveling waves. In this problem, we have:\[y_s(x, t) = y_1(x, t) + y_2(x, t) = (6.00 \text{ mm}) \sin(4.00 \pi x - 400 \pi t) + (6.00 \text{ mm}) \sin(4.00 \pi x + 400 \pi t)\]Using the trigonometric identity for adding sine functions, the standing wave equation becomes:\[y_s(x, t) = (12.00 \text{ mm}) \sin(4.00 \pi x) \cos(400 \pi t)\]
02

Identify the Antinodes and Nodes

In a standing wave, the antinodes occur where the sine component \(\sin(4.00\pi x)\) is at its maximum value (i.e., \(|\sin(4.00\pi x)| = 1\)). This happens at:\[4\pi x = \left(n + \frac{1}{2}\right)\pi, \quad n = 0, 1, 2, \ldots \]Solving for \(x\), the antinodes are located at:\[x = \left(n + \frac{1}{2}\right)\frac{1}{4}\, \text{m}\]
03

Determine the Time for Maximum Displacement Change

The antinode moves from maximum upward displacement to maximum downward displacement as the cosine component \(\cos(400\pi t)\) changes from 1 to -1. This happens over a half period of the cosine function.The period \(T\) of \(\cos(400\pi t)\) is given by:\[T = \frac{2\pi}{400\pi} = 0.005\, \text{s}\]Half of this period is:\[\frac{T}{2} = 0.0025\, \text{s}\]
04

Calculate the Distance Each Wave Moves

During the time interval \(0.0025\, \text{s}\), the distance each wave moves can be calculated using the wave speed. The wave speed \(v\) is given by \(v = \frac{\text{wavelength}\ \lambda}{T}\), where \(\lambda = \frac{2\pi}{k}\) with \(k = 4\pi\).\[\lambda = \frac{2\pi}{4\pi} = 0.5\, \text{m}\]\[v = f\lambda = \frac{1}{0.005} \times 0.5 = 100\, \text{m/s}\]Thus, the distance moved by each wave is:\[\text{distance} = v \times \frac{T}{2} = 100 \times 0.0025 = 0.25\, \text{m}\]

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Key Concepts

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

Wave Interference
When two or more waves meet while traveling along the same medium, they interfere with each other. This can occur in several ways, but two primary types are known as constructive and destructive interference. In constructive interference, the amplitudes of the waves add together, creating a larger wave. Conversely, in destructive interference, the waves cancel each other out, resulting in a reduced or completely nullified wave.

Standing waves are a special scenario of wave interference. They occur when two waves of the same frequency and amplitude travel in opposite directions and interfere. Instead of appearing as separate traveling waves, a standing wave appears to stand still, with certain points called nodes (points of no displacement) and antinodes (points of maximum displacement) staying fixed along the medium.

In the given problem, the standing wave is formed by the addition of two waves traveling in opposite directions. By using trigonometric identities, the interference pattern creates a sinusoidal standing wave where nodes and antinodes can be clearly identified.
Wave Velocity
Wave velocity is the speed at which a wave travels through a medium. It's an essential aspect to understand the movement of waves, particularly in problems related to standing waves. The velocity can be calculated using the formula:
  • v = f\(\lambda\)
in which \(v\) represents wave velocity, \(f\) is the frequency, and \(\lambda\) (lambda) is the wavelength of the wave.

In the context of our exercise, with the given wave equation, the angular wave number \(k\) can be extracted, allowing the calculation of the wavelength \(\lambda\) using \(\lambda = \frac{2\pi}{k}\). This gives us a wavelength of 0.5 meters. Given the periodic nature of the system, we can determine the wave speed as 100 m/s.

Wave velocity is crucial for solving how far each wave travels over a certain time interval. In half of a wave period (0.0025 seconds in this scenario), the waves in the exercise move 0.25 meters along the string. Always remember that knowing the wave speed helps in predicting how quickly wave energies are transferred through a medium.
Antinodes and Nodes
In standing waves, antinodes and nodes are fundamental concepts that help describe wave behavior. Nodes are points along the wave that experience no displacement. Mathematically, they occur where the wave interference results in a consistent zero amplitude, effectively meaning the medium at this point does not move at all.

Antinodes, on the other hand, are the points of maximum displacement within a standing wave. These are located midway between nodes, where the wave interference has maximum constructive interference, resulting in the highest amplitude.

In our exercise, we identify antinodes through the equation \(4\pi x = \left(n + \frac{1}{2}\right)\pi\). Solving for \(x\) gives us the positions where all antinodes are located. These are found at intervals of \(\left(n + \frac{1}{2}\right)\frac{1}{4}\) meters, confirming that antinodes repeat regularly along the length of the string.

Understanding the locations of antinodes and nodes is vital for correctly analyzing wave patterns, especially in practical applications such as sound or light wave phenomena, where such patterns can disrupt or enhance signals based on their position relative to barriers or interfaces.

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Most popular questions from this chapter

A sinusoidal transverse wave of amplitude \(y_{m}\) and wavelength \(\lambda\) travels on a stretched cord. (a) Find the ratio of the maximum particle speed (the speed with which a single particle in the cord moves transverse to the wave) to the wave speed. (b) Does this ratio depend on the material of which the cord is made?

A generator at one end of a very long string creates a wave given by $$ y=(6.0 \mathrm{~cm}) \cos \frac{\pi}{2}\left[\left(2.00 \mathrm{~m}^{-1}\right) x+\left(8.00 \mathrm{~s}^{-1}\right) t\right] $$ and a generator at the other end creates the wave $$ y=(6.0 \mathrm{~cm}) \cos \frac{\pi}{2}\left[\left(2.00 \mathrm{~m}^{-1}\right) x-\left(8.00 \mathrm{~s}^{-1}\right) t\right] $$ Calculate the (a) frequency, (b) wavelength, and (c) speed of each wave. For \(x \geq 0\), what is the location of the node having the (d) smallest, (e) second smallest, and (f) third smallest value of \(x\) ? For \(x \geq 0\), what is the location of the antinode having the (g) smallest, (h) second smallest, and (i) third smallest value of \(x\) ?

A \(125 \mathrm{~cm}\) length of string has mass \(2.00 \mathrm{~g}\) and tension \(7.00 \mathrm{~N}\). (a) What is the wave speed for this string? (b) What is the lowest resonant frequency of this string?

A string under tension \(\tau_{i}\) oscillates in the third harmonic at frequency \(f_{3}\), and the waves on the string have wavelength \(\lambda_{3}\). If the tension is increased to \(\tau_{f}=4 \tau_{i}\) and the string is again made to oscillate in the third harmonic, what then are (a) the frequency of oscillation in terms of \(f_{3}\) and (b) the wavelength of the waves in terms of \(\lambda_{3} ?\)

The speed of a transverse wave on a string is \(170 \mathrm{~m} / \mathrm{s}\) when the string tension is \(120 \mathrm{~N}\). To what value must the tension be changed to raise the wave speed to \(180 \mathrm{~m} / \mathrm{s}\) ?

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