/*! 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 25 A wave has the following propert... [FREE SOLUTION] | 91Ó°ÊÓ

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A wave has the following properties: amplitude \(=0.37 \mathrm{~m}\), period \(=0.77 \mathrm{~s}\), wave speed \(=12 \mathrm{~m} / \mathrm{s} .\) The wave is traveling in the \(-x\) direction. What is the mathematical expression (similar to Equation 16.3 or 16.4 ) for the wave?

Short Answer

Expert verified
\(\psi(x, t) = 0.37 \sin(0.68x + 8.16t)\) is the wave equation.

Step by step solution

01

Identify the formula structure

The general form for a wave traveling in the negative x-direction can be expressed as:\[ \psi(x, t) = A \sin(kx + \omega t + \phi) \]. Here, \(A\) is the amplitude, \(k\) is the wave number, \(\omega\) is the angular frequency, and \(\phi\) is the phase constant.
02

Find angular frequency

The angular frequency \(\omega\) is related to the period \(T\) by the formula: \(\omega = \frac{2\pi}{T}\). Given that the period \(T = 0.77\, \mathrm{s}\), we calculate \(\omega\) as follows:\[ \omega = \frac{2\pi}{0.77} \approx 8.16\, \mathrm{rad/s} \].
03

Find wave number

The wave number \(k\) is related to the wavelength \(\lambda\) via the formula: \(k = \frac{2\pi}{\lambda}\). Since wave speed \(v\) is given by \(v = \lambda f\) and frequency \(f = \frac{1}{T}\), first calculate the wavelength:\[ \lambda = \frac{v}{f} = \frac{12\, \mathrm{m/s}}{\frac{1}{0.77\, \mathrm{s}}} = 12 \times 0.77 = 9.24\, \mathrm{m} \].\Then, calculate \(k\):\[ k = \frac{2\pi}{9.24} \approx 0.68\, \mathrm{rad/m} \].
04

Write the wave equation

Now, with values calculated, plug them into the wave equation formula. The wave equation is now:\[ \psi(x, t) = 0.37 \sin(0.68x + 8.16t) \]. Since the wave travels in the negative x-direction, maintain the plus sign with \(x\) term.

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Key Concepts

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

Wave Properties
Waves are traveling disturbances that carry energy from one place to another without the accompanying transport of matter. Several key properties define a wave: amplitude, wavelength, frequency, period, and wave speed. Understanding these properties helps us describe and analyze wave behavior.

- **Amplitude** refers to the maximum displacement of the wave from its rest position. It determines the wave's intensity.
- **Wavelength** (\( \lambda \)) is the distance between successive points of similar phase on the wave, such as crest to crest or trough to trough.
- **Frequency** (\( f \)) is the number of cycles the wave completes in one second and is measured in hertz (Hz).
- **Period** (\( T \)) is the time it takes for one complete cycle of a wave to pass a given point.
- **Wave speed** (\( v \)) is the rate at which the wave propagates through a medium and is calculated by the product of frequency and wavelength: \( v = f \lambda \).

These basic properties remain consistent for all forms of waves, whether they're sound, light, or water waves.
Wave Number
The wave number (\( k \)) is a crucial concept when discussing wave mathematics. It is defined as the number of wavelengths per unit distance and is given by the formula:\[ k = \frac{2\pi}{\lambda} \]where \( \lambda \) is the wavelength.

The wave number indicates how many wavelengths fit into a given length and is measured in radians per meter (rad/m). By providing insights into the spatial frequency of the wave, the wave number helps describe the wave's structure and is pivotal for solving wave equations. In the context of the given exercise, the wave number gives us a direct understanding of how tight or spread apart the wave oscillations are over a particular distance.
Angular Frequency
Angular frequency (\( \omega \)) is a measure of how quickly the phase of the wave changes with time. It is related to the period (\( T \)) of the wave by the equation:\[ \omega = \frac{2\pi}{T} \]where \( T \) is the period.

Typically measured in radians per second (rad/s), angular frequency tells us how many radians the wave travels through per unit of time. This concept is essential when considering cyclical processes, as it provides an understanding of how fast the wave's oscillations occur over time.

In our exercise, angular frequency is calculated based on the given period, allowing us to express the wave's speed in terms of how rapidly a point on the wave oscillates up and down.
Amplitude Calculation
Amplitude calculation involves determining the maximum extent of oscillation of the wave from its equilibrium position. This parameter is crucial because it dictates the energy carried by the wave. In mathematical terms, the amplitude \( A \) appears as a coefficient in the wave equation:\[ \psi(x, t) = A \sin(kx + \omega t + \phi) \]where \( A \) represents the amplitude.

For our specific exercise, the amplitude is given as 0.37 meters. This value directly influences the intensity of the wave's oscillations, and it is integral to constructing the complete wave equation. Accurately calculating and substituting amplitude into the wave equation helps describe the wave's behavior visually and physically.

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Most popular questions from this chapter

Concept Questions Example 4 in the text discusses an ultrasonic ruler that displays the distance between the ruler and an object, such as a wall. The ruler sends out a pulse of ultrasonic sound and measures the time it takes for the pulse to reflect from the object and return. The ruler uses this time, along with a preset value for the speed of sound in air, to determine the distance. Suppose you use this ruler underwater, rather than in air. (a) Is the speed of sound in water greater than, less than, or equal to the speed of sound in air? (b) Is the reading on the ruler greater than, less than, or equal to the actual distance? Provide reasons for your answers. Problem The actual distance from the ultrasonic ruler to an object is \(25.0 \mathrm{~m} .\) The adiabatic bulk modulus and density of seawater are \(B_{\mathrm{ad}}=2.37 \times 10^{9} \mathrm{~Pa}\) and \(\rho=1025\) \(\mathrm{kg} / \mathrm{m}^{3},\) respectively. Assume that the ruler uses a preset value of \(343 \mathrm{~m} / \mathrm{s}\) for the speed of sound in air, and determine the distance reading on its display. Verify that your answer is consistent with your answers to the Concept Questions.

The middle \(C\) string on a piano is under a tension of \(944 \mathrm{~N}\). The period and wavelength of a wave on this string are \(3.82 \mathrm{~ms}\) and \(1.26 \mathrm{~m}\), respectively. Find the linear density of the string.

Concept Questions Using \(1 \times 10^{-12} \mathrm{~W} / \mathrm{m}^{2}\) as a reference intensity, the threshold of hearing for an average young person is \(0 \mathrm{~dB}\). Two individuals who are not average have thresholds of hearing that are \(\beta_{1}=-8.00 \mathrm{~dB}\) and \(\beta_{2}=+12.0 \mathrm{~dB} .(\mathrm{a})\) Is the sound intensity at a level of \(-8.00 \mathrm{~dB}\) greater than, or less than, that at \(0 \mathrm{~dB} ?\) (b) Which individual has the better hearing? Why? Problem How many times greater (or less) is the sound intensity when person 1 \(\left(\beta_{1}=-8.00 \mathrm{~dB}\right)\) hears it at the threshold of hearing compared to when person 2 \(\left(\beta_{2}=+12.0 \mathrm{~dB}\right)\) hears it at the threshold of hearing? Check to see that your answer is consistent with your answers to the Concept Questions.

A microphone is attached to a spring that is suspended from the ceiling, as the drawing indicates. Directly below on the floor is a stationary 440 -Hz source of sound. The microphone vibrates up and down in simple harmonic motion with a period of \(2.0 \mathrm{~s}\). The difference between the maximum and minimum sound frequencies detected by the microphone is \(2.1 \mathrm{~Hz}\). Ignoring any reflections of sound in the room and using \(343 \mathrm{~m} / \mathrm{s}\) for the speed of sound, determine the amplitude of the simple harmonic motion.

A speeder looks in his rearview mirror. He notices that a police car has pulled behind him and is matching his speed of \(38 \mathrm{~m} / \mathrm{s}\). The siren on the police car has a frequency of \(860 \mathrm{~Hz}\) when the police car and the listener are stationary. The speed of sound is \(343 \mathrm{~m} / \mathrm{s}\). What frequency does the speeder hear when the siren is turned on in the moving police car?

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