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Chapter 16: Problem 27

A wave traveling in the \(+x\) direction has an amplitude of \(0.35 \mathrm{m}\) a speed of \(5.2 \mathrm{m} / \mathrm{s},\) and a frequency of \(14 \mathrm{Hz} .\) Write the equation of the wave in the form given by either Equation 16.3 or 16.4

Short Answer

Expert verified
The wave equation is \( y(x, t) = 0.35 \sin(16.93 x - 28\pi t) \).

Step by step solution

01

Identify known values

We have been given the amplitude \(A = 0.35 \, \text{m}\), speed \(v = 5.2 \, \text{m/s}\), and frequency \(f = 14 \, \text{Hz}\) of the wave.
02

Use wave equation

A wave traveling in the positive \(x\) direction can be described by the equation: \[ y(x, t) = A \sin(kx - \omega t)\] where \(y(x, t)\) represents the wave function, \(A\) is the amplitude, \(k\) is the wave number, \(\omega\) is the angular frequency, \(x\) is the position, and \(t\) is the time.
03

Calculate the angular frequency \(\omega\)

Angular frequency \(\omega\) is given by \(\omega = 2\pi f\). Thus, \(\omega = 2\pi \times 14 \, \text{Hz} = 28\pi \, \text{rad/s}\).
04

Calculate the wave number \(k\)

The wave number \(k\) can be calculated from \(k = \frac{2\pi}{\lambda}\). The wavelength \(\lambda\) is given by \(\lambda = \frac{v}{f}\). Thus, \(\lambda = \frac{5.2 \, \text{m/s}}{14 \, \text{Hz}} \approx 0.371 \, \text{m}\). Therefore, \(k = \frac{2\pi}{0.371} \approx 16.93 \, \text{m}^{-1}\).
05

Write the wave equation

Substitute the values into the wave equation: \[y(x, t) = 0.35 \sin(16.93 x - 28\pi t)\] This equation describes the wave traveling in the positive \(x\) direction.

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Key Concepts

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

Amplitude
Amplitude is a fundamental feature of a wave, representing the wave's maximum displacement from its rest position. In simpler terms, it's the height of the wave's peaks. This concept is essential because it tells us about the wave's energy. The larger the amplitude, the more energy the wave carries.

In our exercise, the amplitude is given as \(0.35\, \text{m}\). This means that at its highest or lowest point, the wave reaches 0.35 meters above or below its equilibrium position. Understanding amplitude helps in visualizing how "tall" or "strong" a wave is and plays a key role in applications like sound waves, where higher amplitudes result in louder sounds.

  • Energy and Amplitude: More energy typically means a larger amplitude.
  • Effects: In sound, larger amplitude translates to louder volume.
Wave Speed
Wave speed describes how quickly a wave propagates through a medium. It is a crucial property as it influences how fast the wave reaches a certain point. In the given problem, the wave speed is 5.2 \(\text{m/s}\). This means the wave advances by 5.2 meters every second.

Wave speed depends on the medium through which the wave is traveling. For example, sound waves travel faster in water than in air, while light waves move slower in glass than in a vacuum. Understanding wave speed is key in various fields such as telecommunications and acoustics.

  • Formula: Wave speed \(v\) is related to frequency \(f\) and wavelength \(\lambda\) by \(v = f\lambda\).
  • Media Dependence: Different materials affect the speed of wave travel.
Angular Frequency
Angular frequency indicates how rapidly a wave oscillates. It is a measure of how many cycles a wave completes in a given period and is expressed in radians per second \(\text{rad/s}\). It is a key concept when dealing with sinusoidal waves like sound or light.

For a wave with a frequency of 14 Hz, the angular frequency \(\omega\) is calculated using the formula \(\omega = 2\pi f\). In our example, \(\omega = 2\pi \times 14 = 28\pi\, \text{rad/s}\). This tells us how fast the wave's phase changes over time.

  • Relation to Frequency: Angular frequency is derived directly from frequency using \(\omega = 2\pi f\).
  • Importance: It helps in analyzing circuits in physics and understanding the behavior of waves in various media.

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Most popular questions from this chapter

A rocket, starting from rest, travels straight up with an acceleration of \(58.0 \mathrm{m} / \mathrm{s}^{2} .\) When the rocket is at a height of \(562 \mathrm{m},\) it produces sound that eventually reaches a ground-based monitoring station directly below. The sound is emitted uniformly in all directions. The monitoring station measures a sound intensity \(I\). Later, the station measures an intensity \(\frac{1}{3} I .\) There are no reflections. Assuming that the speed of sound is \(343 \mathrm{m} / \mathrm{s}\) find the time that has elapsed between the two measurements.

As a prank, someone drops a water-filled balloon out of a window. The balloon is released from rest at a height of \(10.0 \mathrm{m}\) above the ears of a man who is the target. Then, because of a guilty conscience, the prankster shouts a warning after the balloon is released. The warning will do no good, however, if shouted after the balloon reaches a certain point, even if the man could react infinitely quickly. Assuming that the air temperature is \(20^{\circ} \mathrm{C}\) and ignoring the effect of air resistance on the balloon, determine how far above the man's ears this point is.

Dolphins emit clicks of sound for communication and echolocation. A marine biologist is monitoring a dolphin swimming in seawater where the speed of sound is \(1522 \mathrm{m} / \mathrm{s}\). When the dolphin is swimming directly away at \(8.0 \mathrm{m} / \mathrm{s},\) the marine biologist measures the number of clicks occurring per second to be at a frequency of \(2500 \mathrm{Hz}\). What is the difference (in Hz) between this frequency and the number of clicks per second actually emitted by the dolphin?

From a vantage point very close to the track at a stock car race, you hear the sound emitted by a moving car. You detect a frequency that is 0.86 times as small as the frequency emitted by the car when it is stationary. The speed of sound is \(343 \mathrm{m} / \mathrm{s} .\) What is the speed of the car?

Two wires are parallel, and one is directly above the other. Each has a length of \(50.0 \mathrm{m}\) and a mass per unit length of \(0.020 \mathrm{kg} / \mathrm{m} .\) However, the tension in wire \(\mathrm{A}\) is \(6.00 \times 10^{2} \mathrm{N},\) and the tension in wire \(\mathrm{B}\) is \(3.00 \times 10^{2} \mathrm{N} .\) Transverse wave pulses are generated simultaneously, one at the left end of wire \(\mathrm{A}\) and one at the right end of wire \(\mathrm{B}\). The pulses travel toward each other. How much time does it take until the pulses pass each other?
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