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Chapter 16: Problem 23

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 \( y(x,t) = 0.35 \sin(16.91x - 28\pi t) \).

Step by step solution

01

Understand Wave Equation Form

Equation 16.3 or 16.4 generally represents a wave traveling in the +x direction as: \( y(x,t) = A \sin(kx - \omega t) \), where \( A \) is the amplitude, \( k \) is the wave number, and \( \omega \) is the angular frequency.
02

Identify Given Values

From the problem, the amplitude \( A = 0.35 \mathrm{~m} \), speed \( v = 5.2 \mathrm{~m/s} \), and frequency \( f = 14 \mathrm{~Hz} \).
03

Calculate Angular Frequency \( \omega \)

The angular frequency is calculated by the formula \( \omega = 2\pi f \). Therefore, \( \omega = 2\pi \times 14 \mathrm{~Hz} = 28\pi \mathrm{~rad/s} \).
04

Calculate Wave Number \( k \)

The wave number \( k \) is given by the formula \( k = \frac{2\pi}{\lambda} \). Since \( \lambda = \frac{v}{f} \), we have \( \lambda = \frac{5.2}{14} \approx 0.3714 \mathrm{~m} \). Thus, \( k = \frac{2\pi}{0.3714} \approx 16.91 \mathrm{~m^{-1}} \).
05

Write the Wave Equation

Substitute the values into the wave equation form: \( y(x,t) = 0.35 \sin(16.91x - 28\pi t) \).

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Key Concepts

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

Amplitude
The amplitude of a wave is a measure of how "tall" or "strong" the wave is, which indicates the maximum extent of oscillation from its equilibrium position. In simpler terms, it's the distance from the wave's midpoint to its peak or trough. For the wave in our problem, the amplitude is given as 0.35 meters. This means that the highest point of this wave is 0.35 meters above the center line, while the lowest point is 0.35 meters below it.
  • Amplitude is crucial in determining the energy carried by the wave. The larger the amplitude, the more energy it conveys.
  • It is a constant for a particular wave and doesn't depend on frequency or wave speed.
Understanding amplitude helps one grasp the significance of the wave's intensity and the effect it may have during interactions with other media.
Angular Frequency
Angular frequency, represented by the symbol \( \omega \), is key to describing how quickly something oscillates in a wave. Essentially, it tells us how fast the wave makes a complete cycle. You can calculate angular frequency using the formula \( \omega = 2\pi f \), where \( f \) is the frequency of the wave. For this problem, the angular frequency is \( 28\pi \) rad/s.
  • Angular frequency is expressed in radians per second.
  • It connects the wave's frequency with its rotational equivalent in mathematics.
Grasping angular frequency allows us to understand the timing and rate of oscillations, which is fundamental in analyzing wave behaviors.
Wave Number
The wave number, denoted as \( k \), tells us how many waves fit into a certain distance or how "tight" the waves are within a space. It is linked with the wavelength \( \lambda \) via the formula \( k = \frac{2\pi}{\lambda} \). In the exercise, once we calculate the wavelength as approximately 0.3714 meters, \( k \) becomes approximately 16.91 m\(^{-1}\).
  • A higher wave number means more waves fit into a given length, indicating a shorter wavelength.
  • The wave number contributes to the spatial frequency of the wave.
Knowing the wave number is important for understanding the distribution and spacing of waves through a medium.
Wave Speed
Wave speed, symbolized as \( v \), quantifies how fast the wave propagates through the medium. It links frequency and wavelength via the equation \( v = f \lambda \). For our wave, the speed is given as 5.2 meters per second.
  • Wave speed indicates how quickly energy is transmitted from one location to another by the wave.
  • A constant wave speed in a given medium reflects a consistent medium and source characteristics.
Comprehending wave speed is essential for predicting how long it will take a wave to travel between two points and for interpreting the nature of the medium through which it travels.
Frequency
Frequency, indicated by the symbol \( f \), determines how often the wave completes a full cycle within a second. This is expressed in hertz (Hz), with our wave having a frequency of 14 Hz. Essentially, this indicates that 14 complete cycles of the wave occur every second.
  • Higher frequencies mean more cycles per second, usually implying more energy.
  • Frequency remains constant for a wave type in the same medium.
Understanding frequency is crucial for applications ranging from sound and light to radio waves, as it directly influences both the behavior and the energy of the wave.

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Most popular questions from this chapter

As the drawing shows, one microphone is located at the origin, and a second microphone is located on the \(+y\) axis. The microphones are separated by a distance of \(D=1.50 \mathrm{~m}\). A source of sound is located on the \(+x\) axis, its distances from microphones 1 and 2 being \(L_{1}\) and \(L_{2}\), respectively. The speed of sound is \(343 \mathrm{~m} / \mathrm{s}\). The sound reaches microphone 1 first, and then, 1.46 ms later, it reaches microphone 2 . Find the distances \(L_{1}\) and \(L_{2}\)

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 \(A\) and one at the right end of wire B. The pulses travel toward each other. How much time does it take until the pulses pass each other?

When an earthquake occurs, two types of sound waves are generated and travel through the earth. The primary, or \(\mathrm{P}\), wave has a speed of about \(8.0 \mathrm{~km} / \mathrm{s}\) and the secondary, or \(\mathrm{S}\), wave has a speed of about \(4.5 \mathrm{~km} / \mathrm{s}\). A seismograph, located some distance away, records the arrival of the \(\mathrm{P}\) wave and then, 78 s later, records the arrival of the S wave. Assuming that the waves travel in a straight line, how far is the seismograph from the earthquake?

You are riding your bicycle directly away from a stationary source of sound and hear a frequency that is \(1.0 \%\) lower than the emitted frequency. The speed of sound is \(343 \mathrm{~m} / \mathrm{s}\). What is your speed?

Multiple-Concept Example 4 presents one method for modeling this type of problem. Civil engineers use a transit theodolite when surveying. A modern version of this device determines distance by measuring the time required for an ultrasonic pulse to reach a target, reflect from it, and return. Effectively, such a theodolite is calibrated properly when it is programmed with the speed of sound appropriate for the ambient air temperature. (a) Suppose the round-trip time for the pulse is \(0.580 \mathrm{~s}\) on a day when the air temperature is \(293 \mathrm{~K},\) the temperature for which the instrument is calibrated. How far is the target from the theodolite? (b) Assume that air behaves as an ideal gas. If the air temperature were \(298 \mathrm{~K}\), rather than the calibration temperature of \(293 \mathrm{~K}\), what percentage error would there be in the distance measured by the theodolite?
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