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Chapter 12: Problem 69

For the stationary wave: \(y=4 \sin (\pi \mathrm{x} / 15) \cos (96 \pi \mathrm{t})\) the distance between a node and the next antinode is (a) \(7.5\) (b) 15 (c) \(22.5\) (d) 30

Short Answer

Expert verified
The distance between a node and the next antinode is 7.5 units (option a).

Step by step solution

01

Understand the Wave Equation

The given equation is for a stationary wave: \(y = 4 \sin (\pi x / 15) \cos (96 \pi t)\). The general form of a stationary wave is \(y = A \sin(kx) \cos(\omega t)\) where \(k\) is the wave number and \(\omega\) is the angular frequency.
02

Determine the Wave Number

From the equation, \(k = \pi/15\). The wave number \(k\) is related to the wavelength \(\lambda\) by the equation \(k = 2\pi / \lambda\). Solve for \(\lambda\): \(\lambda = 2\pi / k = 2\pi / (\pi / 15) = 30\). This means the full wavelength of the wave is 30 units.
03

Calculate Node to Antinode Distance

The distance between a node and the next antinode is a quarter of the wavelength (\(\lambda/4\)). Since we found \(\lambda = 30\), we calculate \(\lambda/4 = 30/4 = 7.5\).
04

Identify the Correct Option

The distance calculated \(7.5\) corresponds to option (a). Thus, the distance between a node and the next antinode in the given stationary wave is 7.5.

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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 representation that describes the behavior of waves. In this particular case, the wave equation for a stationary wave is given by \(y=4 \sin(\pi x / 15) \cos(96 \pi t)\). Stationary waves are unique because they do not transport energy in the direction of wave propagation. Instead, they appear to vibrate in place.
  • The equation contains two functions: a sine function and a cosine function.
  • The sine function \(\sin(kx)\) relates to the spatial aspect of the wave, where \(k\) is the wave number.
  • The cosine function \(\cos(\omega t)\) is linked to the time-dependent behavior, where \(\omega\) is the angular frequency.
The general form helps us identify important features such as wavelength and wave number that are deeply tied to the wave's physical characteristics. Understanding this equation is foundational to solving problems involving stationary waves.
Wave Number
The wave number, denoted by \(k\), is an essential component in understanding wave behavior. It is the spatial frequency of the wave, meaning how often the wave repeats within a unit of distance. In our given wave equation, the wave number \(k\) has been determined as \(\pi/15\). This indicates the rate of spatial repetition for the sine component of the stationary wave. Here's how to interpret the wave number:
  • A larger wave number means more wave cycles fit into a given distance.
  • A smaller wave number means fewer cycles within the same distance.
  • The relationship between wave number and wavelength is reciprocal; as one increases, the other decreases.
Understanding the wave number's role helps in calculating other crucial properties of the wave, such as wavelength.
Wavelength
Wavelength, represented by \(\lambda\), is the distance over which the wave's shape repeats. It is one of the fundamental characteristics of a wave and is inversely related to the wave number. In the context of our exercise, we learned that the wavelength \(\lambda\) is calculated using the formula \(\lambda = 2\pi / k\). Given \(k = \pi/15\), we find that the wavelength \(\lambda\) is 30 units. Key points about wavelength:
  • It determines the scale of the wave along the axis where the wave is propagating.
  • Knowing the wavelength helps us to calculate distances between nodes and antinodes.
  • Wavelength is essential for understanding the physical dimensions and vibrations of a wave.
Recognizing the relationship between wavelength and the spatial distribution of a wave is crucial for various applications in wave physics.
Nodes and Antinodes
Nodes and antinodes are distinctive features in the formation of stationary waves. They result from the superposition of two waves traveling in opposite directions.
  • Nodes are points of destructive interference where the wave amplitude is consistently zero.
  • Antinodes occur where constructive interference leads to maximum wave amplitude.
  • The distance between consecutive nodes or antinodes is half a wavelength.
  • The distance between a node and an adjacent antinode is a quarter of the wavelength.
In our example, with a wavelength \(\lambda = 30\), the distance from a node to the next antinode is \(7.5\) units, as calculated by \(\lambda/4\). Recognizing these features allows us to predict the stationary wave鈥檚 behavior at different points and is instrumental for problems related to wave interference and resonance.

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Most popular questions from this chapter

Two vibrating tuning forks producing progressive waves given by: \(y_{1}=4 \sin (500 \pi t)\) and \(y_{2}=2 \sin (506 \pi t)\) are held near the ear of a person. If the number of beats heard per second be \(B\) and the ratio of maximum to minimum intensity be \(A\), then (a) \(B=3\) and \(A=2\) (b) \(B=3\) and \(A=9\) (c) \(B=6\) and \(A=2\) (d) \(B=6\) and \(\mathrm{A}=9\)

A sound wave of wavelength \(90 \mathrm{~cm}\) in glass is refracted into air. If the velocity of sound in glass is \(5400 \mathrm{~m} / \mathrm{sec}\), the wavelength of the wave in air is (a) \(55 \mathrm{~cm}\) (b) \(5.5 \mathrm{~cm}\)

A blast gives a sound of intensity \(0.8 \mathrm{~W} / \mathrm{m}^{2}\) and frequency \(1 \mathrm{kHz}\). If the density of air is \(1.3 \mathrm{~kg} / \mathrm{m}^{3}\) and speed of sound in air is \(330 \mathrm{~m} / \mathrm{s}\), the amplitude of the sound wave is approximately (a) \(5 \times 10^{-6} \mathrm{~m}\) (b) \(9.7 \times 10^{-6} \mathrm{~m}\) (c) \(15 \times 10^{-6} \mathrm{~m}\) (d) \(20 \times 10^{6} \mathrm{~m}\)

In a stationary wave (a) energy is uniformly distributed (b) energy is maximum at nodes and minimum at antinodes (c) energy is minimum at nodes and maximum at antinodes (d) alternating maxima and minima of energy are produced at nodes and antinodes

When two waves of almost equal frequencies \(n_{1}\) and \(n_{2}\) are produced simultaneously, then the time interval between successive maxima is (a) \(\frac{1}{n_{1}-n_{2}}\) (b) \(\frac{1}{n_{1}}-\frac{1}{n_{2}}\) (c) \(\frac{1}{n_{1}}+\frac{1}{n_{2}}\) (d) \(\frac{1}{n_{1}+n_{2}}\)
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