/*! 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 6 A certain transverse wave is des... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 15: Problem 6

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
The wave's amplitude is 6.50 mm, wavelength is 28.0 cm, frequency is 27.78 Hz, speed is 7.78 m/s, and it propagates in the positive x-direction.

Step by step solution

01

Identify the Amplitude

The amplitude is the coefficient in front of the cosine function in the wave equation. For the given wave, this coefficient is 6.50 mm. Therefore, the amplitude is 6.50 mm.
02

Determine the Wavelength

The wave equation is given in the form \[ y(x, t) = A \cos(2\pi \left(\frac{x}{\lambda} - \frac{t}{T}\right)) \] where \(\lambda\) is the wavelength. Comparing with the given equation, \[ y(x, t) = (6.50 \text{ mm}) \cos\left(2\pi\left(\frac{x}{28.0 \text{ cm}} - \frac{t}{0.0360 \text{ s}}\right)\right) \]we find \(\lambda = 28.0 \text{ cm}\). Thus, the wavelength is 28.0 cm.
03

Calculate the Frequency

The frequency is given by the reciprocal of the period \(T\). Here, the equation is in the form \[ \frac{t}{T} = \frac{t}{0.0360 \text{ s}} \]which implies \(T = 0.0360 \text{ s}\). Therefore, the frequency \(f = \frac{1}{T} = \frac{1}{0.0360} \approx 27.78 \text{ Hz}\).
04

Find the Speed of Propagation

The speed \(v\) of the wave can be determined using the formula \[ v = \lambda f \]where \(\lambda = 28.0 \text{ cm} = 0.28 \text{ m}\) and \(f = 27.78 \text{ Hz}\). Thus,\[ v = 0.28 \text{ m} \times 27.78 \text{ Hz} \approx 7.78 \text{ m/s} \].
05

Determine the Direction of Propagation

Since the wave equation has the form \( \cos(2\pi (\frac{x}{\lambda} - \frac{t}{T})) \), the wave is propagating in the positive x-direction. This is because the negative sign in the argument \( -\frac{t}{T} \) indicates movement in the 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 determination
The amplitude of a wave is a measure of its maximum displacement from the rest position. Think of it as the height of the wave from the center line to the peak. In mathematical terms, it is the coefficient that precedes the cosine or sine function in the wave equation. For the given transverse wave described by the equation \( y(x, t) = (6.50 \text{ mm}) \cos 2\pi\left(\frac{x}{28.0 \text{ cm}} - \frac{t}{0.0360 \text{ s}}\right) \),
  • the amplitude is \( 6.50 \text{ mm} \).
This means that the wave will rise or fall 6.50 mm from the center line in any direction. Knowing the amplitude is useful to understand the energy the wave carries: a larger amplitude means a stronger wave.
Wavelength calculation
Wavelength is the distance between two consecutive points of a wave that are in phase, such as the distance from one peak to the next. In wave equations, wavelength is represented by \( \lambda \). It shapes the wave's spatial periodicity, meaning how often waves repeat per unit distance.

In the provided wave equation, comparing with the generic form \( y(x, t) = A \cos\left(2\pi\left(\frac{x}{\lambda} - \frac{t}{T}\right)\right) \),
  • the term \( \frac{x}{28.0 \text{ cm}} \) tells us that the wavelength \( \lambda \) is \( 28.0 \text{ cm} \).
The wavelength helps in understanding the scale of a wave and how far apart its features, like peaks and troughs, are along the x-direction.
Frequency calculation
Frequency refers to how often the wave oscillates through its cycle in a given period. It is calculated using the period \( T \), which is the time it takes to complete one cycle of oscillation. Frequency \( f \) is thus the reciprocal of the period:
  • \( f = \frac{1}{T} \).
For this wave, the equation element \( \frac{t}{0.0360 \text{ s}} \) indicates that \( T = 0.0360 \text{ s} \), leading to a frequency of \( 27.78 \text{ Hz} \). Frequency is key to understanding how quickly wave events occur. A higher frequency means more cycles per second, often corresponding to sounds we perceive as higher pitched.
Wave speed
Wave speed represents how fast the wave travels through a medium. It is an essential characteristic that combines both the spatial and temporal aspects of wave motion. Mathematically, it is given by the product of wavelength and frequency:
  • \( v = \lambda f \).
In this example, with a wavelength of \( 28.0 \text{ cm} \) converted to meters \( \lambda = 0.28 \text{ m} \) and frequency \( f = 27.78 \text{ Hz} \),
  • the wave speed \( v \approx 7.78 \text{ m/s} \).
Understanding wave speed is crucial in contexts like communication or seismology, indicating how fast information or energy propagates through the medium.
Wave propagation direction
The direction in which a wave travels is indicated by the phase part of the wave equation, especially the signs in its formula. In the wave equation \( \cos(2\pi (\frac{x}{\lambda} - \frac{t}{T})) \), the term \( -\frac{t}{T} \) indicates that the wave moves in the positive x-direction.

Think about it like this: as time increases, the effect must offset the position \( x \) component in the equation to maintain steady phase conditions. Hence,
  • the negative sign before the time term \(-\frac{t}{T}\) suggests that the wave is propagating in the positive x-direction.
This knowledge is vital when analyzing scenarios like signal transmission, where directionality dictates the flow of information.

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

Wave Equation and Standing Waves. (a) Prove by direct substitution that \(y(x, t)=\left(A_{\text { sw }} \sin k x\right)\) sin \(\omega t\) is a solution of the wave equation, Eq. \((15.12),\) for \(v=\omega / k .\) (b) Explain why the relationship \(v=\omega / k\) for traveling waves also applies to standing waves.

Energy in a Triangular Pulse. A triangular wave pulse on a taut string travels in the positive \(x\) -direction with speed \(v\) . The tension in the string is \(F,\) and the linear mass density of the string is \(\mu .\) At \(t=0,\) the shape of the pulse is given by $$ y(x, 0)=\left\\{\begin{array}{ll}{0} & {\text { if } x<-L} \\ {h(L+x) / L} & {\text { for }-LL}\end{array}\right. $$ (a) Draw the pulse at \(t=0 .\) (b) Determine the wave function \(y(x, t)\) at all times \(t\) (c) Find the instantancous power in the wave. Show that the power is zero except for \(-L<(x-v t)

Audible Sound. Provided the amplitude is sufficiently great, the human ear can respond to longitudinal waves over a range of frequencies from about 20.0 \(\mathrm{Hz}\) to about 20.0 \(\mathrm{kHz}\) . (a) If you were to mark the beginning of each complete wave pattern with a red dot for the long-wavelength sound and a blue dot for the short-wavelength sound, how far apart wonld the red dots be, and how far apart would the blue dots be? (b) In reality would adjacent dots in each set be far enough apart for you to easily measure their separation with a meterstick? (c) Suppose you repeated part (a) in water, where sound travels at 1480 \(\mathrm{m} / \mathrm{s}\) . How far apart would the dots be in each set? Could you readily measure their separation with a meterstick?

A sinusoidal transverse wave travels on a string. The string has length 8.00 \(\mathrm{m}\) and mass 6.00 \(\mathrm{g}\) . The wave speed is 30.0 \(\mathrm{m} / \mathrm{s}\) , and the wavelength is 0.200 \(\mathrm{m}\) (a) If the wave is to have an aver- age power of 50.0 \(\mathrm{W}\) , what must be the amplitude of the wave? (b) For this same string, if the amplitude and wavelength are the same as in part (a), what is the average power for the wave if the tension is increased such that the wave speed is doubled?

A wire with mass 40.0 \(\mathrm{g}\) is stretched so that its ends are tied down at points 80.0 \(\mathrm{cm}\) apart. The wire vibrates in its fundamental mode with frequency 60.0 \(\mathrm{Hz}\) and with an amplitude at the antinodes of 0.300 \(\mathrm{cm}\) . (a) What is the speed of propagation of transverse waves in the wire? (b) Compute the tension in the wire. (c) Find the maximum transverse velocity and acceleration of particles in the wire.
See all solutions

Recommended explanations on Physics Textbooks

Physics of Motion

Read Explanation

Scientific Method Physics

Read Explanation

Rotational Dynamics

Read Explanation

Wave Optics

Read Explanation

Further Mechanics and Thermal Physics

Read Explanation

Linear Momentum

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.