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Chapter 14: Problem 19

Write an expression for a harmonic wave with an amplitude of \(0.16 \mathrm{m},\) a wavelength of \(2.1 \mathrm{m},\) and a period of \(1.8 \mathrm{s}\). The wave is transverse, travels to the right, and has a displacement of \(0.16 \mathrm{m}\) at \(t=0\) and \(x=0\)

Short Answer

Expert verified
The wave equation is \( y(x, t) = 0.16 \cos(2.99x - 3.49t) \).

Step by step solution

01

Understand the given parameters

We are given the amplitude \( A = 0.16 \, \text{m} \), wavelength \( \lambda = 2.1 \, \text{m} \), and period \( T = 1.8 \, \text{s} \) of a harmonic wave that travels to the right.
02

Determine phase constant

Since the wave is transverse and has a displacement of \( 0.16 \, \text{m} \) at \( t = 0 \) and \( x = 0 \), the initial displacement is equal to the amplitude. This means the phase constant \( \phi \) for a cosine wave is \( 0 \) radians because \( \cos(0) = 1 \).
03

Calculate wave number

The wave number \( k \) is related to the wavelength by \( k = \frac{2\pi}{\lambda} \). Substitute \( \lambda = 2.1 \): \[ k = \frac{2\pi}{2.1} \approx 2.99 \, \text{m}^{-1} \].
04

Calculate angular frequency

The angular frequency \( \omega \) is related to the period by \( \omega = \frac{2\pi}{T} \). Substitute \( T = 1.8 \): \[ \omega = \frac{2\pi}{1.8} \approx 3.49 \, \text{s}^{-1} \].
05

Write the wave equation

The general form of the wave equation is \( y(x, t) = A \cos(kx - \omega t + \phi) \). Substituting in the known values: \[ y(x, t) = 0.16 \cos(2.99x - 3.49t) \].

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Key Concepts

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

Wave Equation
The wave equation is a mathematical formula used to describe the behavior of waves. For harmonic waves, the equation takes the form
  • \( y(x, t) = A \cos(kx - \omega t + \phi) \)
Here,
  • \( y(x, t) \) represents the displacement of the wave at position \( x \) and time \( t \)
  • \( A \) is the amplitude
  • \( k \) is the wave number
  • \( \omega \) is the angular frequency
  • \( \phi \) is the phase constant

The wave equation captures how the wave propagates through space and time. In our exercise, substituting the given values results in the equation: \( y(x, t) = 0.16 \cos(2.99x - 3.49t) \). This describes a transverse wave moving to the right with the specified amplitude, wave number, and angular frequency.
Wave Number
The wave number \( k \) is a measure of spatial frequency, telling us how many wave cycles fit into a unit length. It is given by the formula:
  • \( k = \frac{2\pi}{\lambda} \)
where \( \lambda \) is the wavelength of the wave.

For our specific case, the wavelength is \( 2.1 \text{ m} \), resulting in a wave number:
  • \( k = \frac{2\pi}{2.1} \approx 2.99 \, \text{m}^{-1} \)
This value signifies the number of complete wave cycles present per meter in the same direction of propagation.
Angular Frequency
Angular frequency \( \omega \) is a measure of how rapidly the wave oscillates in time. It relates directly to the period \( T \) of the wave:
  • \( \omega = \frac{2\pi}{T} \)
By calculating using the given period, \( T = 1.8 \text{ s} \), we find that:
  • \( \omega = \frac{2\pi}{1.8} \approx 3.49 \, \text{s}^{-1} \)


This frequency essentially indicates how many cycles are completed per second. A higher angular frequency means the wave cycles more quickly. In our equation, this value signifies the oscillation speed of the harmonic wave as it moves to the right.
Phase Constant
The phase constant \( \phi \) in the wave equation adjustments the initial phase of the wave. It is often determined by initial conditions, such as displacement at \( t = 0 \) and \( x = 0 \). In the given problem, since the displacement matches the amplitude at these coordinates, we know that \( \cos(\phi) = 1 \).

This indicates an initial phase constant of \( \phi = 0 \) radians.
  • In practice, a phase constant of zero means the wave starts at its maximum displacement at time zero, aligning perfectly with the cosine function
This simplifies our solution, allowing us to focus on the other properties like wave number and angular frequency, without changing the wave's starting point.

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Most popular questions from this chapter

The speed of surface waves in water decreases as the water becomes shallower. Suppose waves travel across the surface of a lake with a speed of \(2.0 \mathrm{m} / \mathrm{s}\) and a wavelength of \(1.5 \mathrm{m}\). When these waves move into a shallower part of the lake, their speed decreases to \(1.6 \mathrm{m} / \mathrm{s}\), though their frequency remains the same. Find the wavelength of the waves in the shallower water.

Sound 1 has an intensity of \(38.0 \mathrm{W} / \mathrm{m}^{2}\). Sound 2 has an intensity level that is \(2.5 \mathrm{dB}\) greater than the intensity level of sound 1\. What is the intensity of sound \(2 ?\)

A surfer floating beyond the breakers notes 14 waves per minute passing her position. If the wavelength of these waves is \(34 \mathrm{m},\) what is their speed?

Two tuning forks have frequencies of \(278 \mathrm{Hz}\) and \(292 \mathrm{Hz}\) What is the beat frequency if both tuning forks are sounded simultancously?

When guitar strings A and B are plucked at the same time, a beat frequency of \(2 \mathrm{Hz}\) is heard. If string \(\mathrm{A}\) is tightened, the beat frequency increases to \(3 \mathrm{Hz}\). Which of the two strings had the lower frequency initially?
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