/*! 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 4 A person driving at \(17 \mathrm... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A person driving at \(17 \mathrm{m} / \mathrm{s}\) crosses the line connecting two radio transmitters at right angles, as shown in Figure \(28-31\). The transmitters emit identical signals in phase with each other, which the driver receives on the car radio. When the car is at point \(A\), the radio picks up a maximum net signal. (a) What is the longest possible wavelength of the radio waves? (b) How long after the car passes point A does the radio experience a minimum in the net signal? Assume that the wavelength has the value found in part (a).

Short Answer

Expert verified
(a) The longest wavelength is twice the distance \(d\) from car to transmitter, \(\lambda = 2d\). (b) Time for minimum signal is \(d/17\) seconds.

Step by step solution

01

Understanding the Problem

The driver is moving across two radio transmitters at a constant speed. The radio in the car receives signals from these transmitters, and at point \(A\), these signals are in phase, resulting in maximum signal reception. Our task is to find the longest possible wavelength (\(\lambda\)) and determine when the signals fall out of phase, causing a minimum in signal reception.
02

Determine Longest Wavelength

For maximum signal at point \(A\), the path length difference between the signals from the transmitters should be a multiple of their wavelength \(\lambda\). The longest wavelength corresponds to the first occurrence, which is when the path difference equals \(\lambda/2\). Thus, \(\lambda = 2d\) where \(d\) is the distance from each transmitter to the car.
03

Find Time for Minimum Signal

For minimum signal reception, the path difference should be \(\lambda/2,\ 3\lambda/2, \ldots\). Considering the longest wavelength, \(\lambda/2 = d\), the first minimum occurs when the path difference is \(\lambda/2\). After moving from point \(A\), the car travels a distance \(\lambda/2\) to the point where the radio experience is minimal.
04

Calculate Time Duration

The speed of the car is given as \(17 \mathrm{m/s}\). The time \(t\) taken to travel a distance \(\lambda/2\) is given by: \[t = \frac{\lambda/2}{17} = \frac{d}{17}.\] Thus, using \(\lambda = 2d\), \[ t = \frac{d}{17}. \]

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.

Radio Waves
Radio waves are a type of electromagnetic wave with long wavelengths that can transmit signals over large distances without the need for physical connections. They are used extensively in everyday technologies such as radios, televisions, and cell phones. These waves have frequencies ranging from 3 kHz to 300 GHz. The ability of radio waves to carry information comes from their capability to modulate, which involves changing some aspect of the wave like amplitude, frequency, or phase to encode data.

Radio transmitters emit radio waves by accelerating electrons to create oscillating electric and magnetic fields. These transmitted signals can travel through various mediums, including air, space, and solid objects. Importantly, radio waves are not absorbed much by air or water, which allows them to travel long distances without significant loss, making them ideal for communication.

When these transmitted signals encounter a receiver like a car radio, they induce an electric signal corresponding to the encoded information. The car's receiver then processes this signal to reproduce the sound originally encoded in the radio waves. This simple transmission and reception process underlies a wide range of communication technologies.
Wavelength
Wavelength is a fundamental property of waves, representing the distance between successive peaks or troughs of a wave. In the context of radio waves, the wavelength is particularly crucial because it determines many characteristics of the wave's propagation and interaction with antennas and receivers.

Wavelength is inversely related to frequency according to the equation \( \lambda = \frac{c}{f} \), where \( \lambda \) is the wavelength, \( c \) is the speed of light, and \( f \) is the frequency. For radio waves, this means longer wavelengths correspond to lower frequencies and vice versa. Long wavelengths allow radio waves to diffract around obstacles more effectively and are typically used for long-distance communication.

In the context of the problem, the longest possible wavelength for maximum signal at point A is determined by the path difference between radio waves from two transmitters. The signals are fully in phase, creating maximum signal strength when the path difference is a multiple of the wavelength. The longest wavelength occurs when this path difference corresponds to a single wavelength, calculated using the geometry of the problem.
Phase Difference
Phase difference refers to the difference in the phase angle between two waves, which affects how these waves interact with each other when they meet. In terms of radio waves, phase difference can cause constructive or destructive interference, leading to areas of strong signal (constructive) or weak signal (destructive).

Constructive interference occurs when the phase difference between two waves is a multiple of \( 2\pi \), meaning the peaks and troughs of one wave align with those of the other. Conversely, destructive interference happens when the phase difference is an odd multiple of \( \pi \), causing the peaks of one wave to align with the troughs of another, effectively canceling each other out.

In the exercise, the car experiences maximum signal at point A because the waves are in phase (constructive interference). As the car moves, the phase difference changes. When the phase difference reaches \( \pi \) or \( 3\pi \), destructive interference occurs, resulting in a minimum signal. Understanding phase difference is key in designing and troubleshooting radio wave systems to ensure reliable communication.

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

(a) Does the path-length difference \(\Delta \ell\) increase or decrease as you move from one bright fringe of a two-slit experiment to the next bright fringe farther out? (b) What is \(\Delta \ell\) in terms of the wavelength \(\lambda\) of the light?

A single-slit diffraction pattern is formed on a distant screen. Assuming the angles involved are small, by what factor will the width of the central bright spot on the screen change if (a) the wavelength is doubled, (b) the slit width is doubled, or (c) the distance from the slit to the screen is doubled?

A lens that is "optically perfect" is still limited by diffraction effects. Suppose a lens has a diameter of \(120 \mathrm{mm}\) and a focal length of \(640 \mathrm{mm}\). (a) Find the angular width (that is, the angle from the bottom to the top) of the central maximum in the diffraction pattern formed by this lens when illuminated with \(540-\mathrm{nm}\) light. (b) What is the linear width (diameter) of the central maximum at the focal distance of the lens?

Early cameras were little more than a box with a pinhole on the side opposite the film. (a) What angular resolution would you expect from a pinhole with a \(0.50-\mathrm{mm}\) diameter? (b) What is the greatest distance from the camera at which two point objects \(15 \mathrm{cm}\) apart can be resolved? (Assume light with a wavelength of \(520 \mathrm{nm} .\) )

A spy camera is said to be able to read the numbers on a car's license plate. If the numbers on the plate are \(5.0 \mathrm{cm}\) apart, and the spy satellite is at an altitude of \(160 \mathrm{km},\) what must be the diameter of the camera's aperture? (Assume light with a wavelength of \(550 \mathrm{nm} .\) )

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.