/*! 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 8 A certain transverse wave is des... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 15: Problem 8

A certain transverse wave is described by $$y(x, t)=(6.50 \mathrm{mm}) \cos 2 \pi\left(\frac{x}{28.0 \mathrm{cm}}-\frac{t}{0.0360 \mathrm{s}}\right)$$ Determine the wave's (a) amplitude; (b) wavelength; (c) frequency; (d) speed of propagation; (e) direction of propagation.

Short Answer

Expert verified
Amplitude: 6.50 mm, Wavelength: 28.0 cm, Frequency: ~27.78 Hz, Speed: ~7.79 m/s, Direction: Positive x-direction.

Step by step solution

01

Identify the Amplitude

The given transverse wave equation is \( y(x, t) = (6.50 \text{ mm}) \cos 2\pi \left( \frac{x}{28.0 \text{ cm}} - \frac{t}{0.0360 \text{ s}} \right) \). The amplitude of the wave is the coefficient of the cosine function, which represents the maximum displacement from the equilibrium position. Here, the amplitude is \(6.50 \text{ mm}\).
02

Determine the Wavelength

In the wave equation, the term \( \frac{x}{28.0 \text{ cm}} \) indicates the spatial part of the wave number. The wavelength \( \lambda \) is associated with the denominator in this term, i.e., \( \lambda = 28.0 \text{ cm} \).
03

Calculate the Frequency

The term \( \frac{t}{0.0360 \text{ s}} \) in the equation reflects the temporal frequency. We know that the period \( T \) of the wave is given by this denominator, so \( T = 0.0360 \text{ s} \). Frequency \( f \) is the reciprocal of the period: \( f = \frac{1}{T} = \frac{1}{0.0360 \text{ s}} \approx 27.78 \text{ Hz} \).
04

Find the Speed of Propagation

The wave speed \( v \) is determined by the relationship \( v = \lambda f \). Substituting the values, \( v = 28.0 \text{ cm} \times 27.78 \text{ Hz} = 778.84 \text{ cm/s} \) or \( v = 7.7884 \text{ m/s} \).
05

Determine the Direction of Propagation

The expression \( \left( \frac{x}{28.0 \text{ cm}} - \frac{t}{0.0360 \text{ s}} \right) \) indicates that the wave travels in the positive x-direction, as the terms maintain a form consistent with \( \left( \frac{x}{\lambda} - \frac{t}{T} \right) \), signaling a positive direction.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Amplitude
The amplitude of a wave refers to the maximum extent of its oscillation or displacement from its mean position. It's like the wave's strength or intensity. Think of it as how high the wave peaks go or how deep the troughs are. In the given wave equation, the amplitude is identified as the number preceding the cosine function. Here, it's 6.50 mm. This means that the wave's highest and lowest points are 6.50 mm above and below the equilibrium position, respectively, which tells you how big the wave is visually. It's an important characteristic because it directly relates to the wave's energy; a larger amplitude indicates a wave carrying more energy.
Wavelength
Wavelength is the distance between identical points of consecutive cycles in a wave, such as peak to peak or trough to trough. This characteristic measures the length of one complete wave cycle in the physical space. It helps to understand how stretched or compact the wave is. In our wave equation, the term \(\frac{x}{28.0 \text{ cm}}\) helps us determine the wavelength, which is 28.0 cm. Wavelength is crucial because it dictates how long the wave takes to 'reset' its pattern over distance. It's important when determining how waves interact with each other and with physical barriers.
Frequency
Frequency refers to the number of cycles a wave completes in one second. It's often measured in hertz (Hz), where one hertz is equivalent to one cycle per second. For our wave, the term involving time, \(\frac{t}{0.0360 \text{ s}}\), allows us to determine the frequency. This section tells us the wave completes one cycle in 0.0360 seconds, leading to a frequency of about 27.78 Hz. Higher frequency means more cycles per second occurring through the medium the wave travels. Frequency is closely linked to the wave's energy and affects how we perceive sound and light in practical applications.
Wave Speed
Wave speed depicts how fast a wave travels through a medium. It's a measure of how quickly the wave propagates as it transports energy from one place to another. The speed can be calculated using the relationship between wavelength and frequency, expressed as \( v = \lambda f \). In this case, with a wavelength of 28.0 cm and a frequency of 27.78 Hz, we compute a wave speed of 778.84 cm/s, or approximately 7.7884 m/s. Wave speed is critical when looking at phenomena like sound waves where speed determines how quickly the sound reaches your ears.
Wave Propagation Direction
The direction of wave propagation indicates the direction in which the wave energy travels. For our transverse wave, the expression \(\left( \frac{x}{28.0 \text{ cm}} - \frac{t}{0.0360 \text{ s}} \right)\) suggests the wave moves in the positive x-direction. This is because the form of the equation, \(\frac{x}{\lambda} - \frac{t}{T}\), is consistent with a wave moving towards positive x-values. Understanding the direction of propagation is essential as it helps predict where and how the wave鈥檚 energy will be delivered, affecting how we use technologies like antennas or sound systems.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

One end of a horizontal rope is attached to a prong of an electrically driven tuning fork that vibrates the rope transversely at 120 \(\mathrm{Hz}\) . The other end passes over a pulley and supports a \(1.50-\mathrm{kg}\) mass. The linear mass density of the rope is 0.0550 \(\mathrm{kg} / \mathrm{m}\) . (a) What is the speed of a transverse wave on the rope? (b) What is the wavelength? (c) How would your answers to parts (a) and (b) change if the mass were increased to 3.00 \(\mathrm{kg}\) ?

A light wire is tightly stretched with tension \(F .\) Trans- verse traveling waves of amplitude \(A\) and wavelength \(\lambda_{1}\) carry average power \(P_{\mathrm{av}, 1}=0.400 \mathrm{W}\) . If the wavelength of the waves is doubled, so \(\lambda_{2}=2 \lambda_{1},\) while the tension \(F\) and amplitude \(A\) are not altered, what then is the average power \(P_{\mathrm{av}, 2}\) carried by the waves?

A transverse sine wave with an amplitude of 2.50 \(\mathrm{mm}\) and a wavelength of 1.80 \(\mathrm{m}\) travels from left to right along a long, horizontal, stretched string with a speed of 36.0 \(\mathrm{m} / \mathrm{s} .\) Take the origin at the left end of the undisturbed string. At time \(t=0\) the left end of the string has its maximum upward displacement. (a) What are the frequency, angular frequency, and wave number of the wave? (b) What is the function \(y(x, t)\) that describes the wave? (c) What is \(y(t)\) for a particle at the left end of the string? (d) What is \(y(t)\) for a particle 1.35 \(\mathrm{m}\) to the right of the origin? (e) What is the maximum magnitude of transverse velocity of any particle of the string? (f) Find the transverse displacement and the transverss velocity of a particle 1.35 \(\mathrm{m}\) to the right of the origin at time \(t=0.0625 \mathrm{s}\)

A large rock that weighs 164.0 \(\mathrm{N}\) is suspended from the lower end of a thin wire that is 3.00 \(\mathrm{m}\) long. The density of the rock is 3200 \(\mathrm{kg} / \mathrm{m}^{3} .\) The mass of the wire is small enough that its effect on the tension in the wire can be neglected. The upper end of the wire is held fixed. When the rock is in air, the fundamental frequency for transverse standing waves on the wire is 42.0 Hz. When the rock is totally submerged in a liquid, with the top of the rock just below the surface, the fundamental frequency for the wire is 28.0 Hz. What is the density of the liquid?

A \(5.00-\mathrm{m}, 0.732-\mathrm{kg}\) wire is used to support two uni-? form \(235-\mathrm{N}\) posts of equal length (Fig. P15.62). Assume that the wire is essentially horizontal and that the speed of sound is 344 \(\mathrm{m} / \mathrm{s} . \mathrm{A}\) strong wind is blowing, causing the wire to vibrate in its 5 th overtone. What are the frequency and wavelength of the sound this wire produces?
See all solutions

Recommended explanations on Physics Textbooks

Torque and Rotational Motion

Read Explanation

Electromagnetism

Read Explanation

Fields in Physics

Read Explanation

Force

Read Explanation

Atoms and Radioactivity

Read Explanation

Further Mechanics and Thermal Physics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.