/*! 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 15 A circular wave travels outward ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A circular wave travels outward from the origin. At one instant of time, the phase at \(r_{1}=20 \mathrm{cm}\) is 0 rad and the phase at \(r_{2}=80 \mathrm{cm}\) is \(3 \pi\) rad. What is the wavelength of the wave?

Short Answer

Expert verified
The wavelength of the wave is 40 cm.

Step by step solution

01

Determine Phase Difference

The phase difference between the two points is given by the difference in their phases. This can be computed as follows: \[\Delta \phi = \phi_{2} - \phi_{1} = 3\pi - 0 = 3\pi \, \text{radians} \]
02

Link to Distance Travelled

The difference in distances from the origin to the two points that correspond to the difference in phases can also be computed. We have \[ r_{2} - r_{1} = 80\, \text{cm} - 20\,\text{cm} = 60 \,\text{cm} \]
03

Compute Wavelength

The phase difference in radian is connected to the distance travelled through the relation \[ \Delta \phi = \frac{2\pi}{\lambda} (r_{2} - r_{1})\] Rearranging for wavelength ‘\(\lambda\)’ gives: \[ \lambda = \frac{2\pi (r_{2} - r_{1})}{\Delta \phi} = \frac{2\pi \times 60}{3\pi} = 40 \,\text{cm} \]

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 91Ó°ÊÓ!

Key Concepts

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

Phase Difference
In wave propagation, understanding phase difference is crucial. Phase difference refers to the amount by which one wave lags or leads another wave. Imagine two ripples spreading out from a point like when you drop two pebbles into a pond.
* The phase is the relative position within one cycle of a wave.* It is usually measured in radians or degrees.* If one wave is at its peak while another is at its trough, they have a phase difference of 180 degrees or \(\pi\) radians.
In the exercise, the wave at \(r_{1} = 20\, \text{cm}\) has a phase of \(0\) radians. At \(r_{2} = 80\, \text{cm}\), the phase is \(3\pi\) radians. This means the phase difference is \(3\pi\) radians, indicating the wave has moved through multiple cycles between these two points.
Recognizing and calculating phase differences help us understand how waves interact and can assist in finding other parameters of the wave.
Wavelength Calculation
A key factor in wave studies is computing the wavelength, which is the physical length of one complete cycle of the wave. We know from the exercise that the phase difference \(\Delta \phi\) is related to the distance and wavelength by:\[ \Delta \phi = \frac{2\pi}{\lambda} (r_{2} - r_{1})\] This equation connects how far the wave travels to the change in its phase.
To find the wavelength \(\lambda\), we rearrange the formula to solve for \(\lambda\):\[\lambda = \frac{2\pi (r_{2} - r_{1})}{\Delta \phi}\]
Substituting the known values:
  • \(r_{2} - r_{1} = 60\, \text{cm}\)
  • \(\Delta \phi = 3\pi\, \text{radians}\)
The calculation becomes:\[\lambda = \frac{2\pi \times 60}{3\pi} = 40 \,\text{cm}\]
Thus, the wavelength of the wave is \(40 \,\text{cm}\). This tells us how far the wave travels along its path for a single cycle to complete.
Circular Wave
Circular waves are fascinating as they propagate outwards in all directions from a central point much like ripples in water. Imagine the symmetry and uniformity of a pond ripple that moves in concentric circles.
* They originate from a point source. * The distance from the point source increases uniformly in all directions. * The waves' fronts are circular arcs that expand over time.
In the exercise we consider the circular wave emanating from the origin. This type of wave propagation helps illustrate how waves behave when spreading out over a surface.
The further out you go from the source, the more potential points you have, each with potentially different phases and amplitudes. Despite these potential changes along the path, by leveraging calculations like phase difference and wavelength, we can predict key wave characteristics. Circular waves are essential for understanding fundamental wave properties and behaviors.

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

Lasers can be used to drill or cut material. One such laser generates a series of high-intensity pulses rather than a continuous beam of light. Each pulse contains \(500 \mathrm{mJ}\) of energy and lasts 10 \(ns\). The laser fires 10 such pulses per second. a. What is the peak power of the laser light? The peak power is the power output during one of the 10 ns pulses. b. What is the average power output of the laser? The average power is the total energy delivered per second. c. Alens focuses the laser beam to a \(10-\mu\) m-diameter circle on the target. During a pulse, what is the light intensity on the target? d. The intensity of sunlight at midday is about \(1100 \mathrm{W} / \mathrm{m}^{2} .\) What is the ratio of the laser intensity on the target to the intensity of the midday sun?

A wave travels with speed \(200 \mathrm{m} / \mathrm{s}\). Its wave number is \(1.5 \mathrm{rad} / \mathrm{m} .\) What are its (a) wavelength and (b) frequency?

A concert loudspeaker suspended high above the ground emits 35 W of sound power. A small microphone with a \(1.0 \mathrm{cm}^{2}\) area is \(50 \mathrm{m}\) from the speaker. a. What is the sound intensity at the position of the microphone? b. How much sound energy impinges on the microphone each second?

The displacement of a traveling wave is $$ D(x, t)=\left\\{\begin{array}{ll} 1 \mathrm{cm} & \text { if }|x-3 t| \leq 1 \\ 0 \mathrm{cm} & \text { if }|x-3 t|>1 \end{array}\right. $$ where \(x\) is in \(m\) and \(t\) in \(s\). a. Draw displacement-versus-position graphs at 1 s intervals from \(t=0\) s to \(t=3 \mathrm{s}\). Use an \(x\) -axis that goes from -2 to \(12 \mathrm{m} .\) Stack the four graphs vertically. b. Determine the wave speed from the graphs. Explain how you did so. c. Determine the wave speed from the equation for \(D(x, t)\) Does it agree with your answer to part b?

A sound wave is described by \(D(y, t)=(0.0200 \mathrm{mm}) \times\) \(\sin [(8.96 \mathrm{rad} / \mathrm{m}) \mathrm{y}+(3140 \mathrm{rad} / \mathrm{s}) t+\pi / 4 \mathrm{rad}],\) where \(\mathrm{y}\) is in \(\mathrm{m}\) and \(t\) is in \(s\) a. In what direction is this wave traveling? b. Along which axis is the air oscillating? c. What are the wavelength, the wave speed, and the period of oscillation? d. Draw a displacement-versus-time graph \(D(y=1.00 \mathrm{m}, t)\) at \(y=1.00 \mathrm{m}\) from \(t=0 \mathrm{s}\) to \(t=4.00 \mathrm{ms}\).

See all solutions

Recommended explanations on Physics Textbooks

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.